In what way do I have to look at this to be able to understand?
The base current increase (decrease) is due to an increase (decrease) in \$v_{BE}\$.
The increase (decrease) in \$v_{BE}\$ increases (decreases) the injection of carriers from the heavily doped emitter.
Most of these carriers cross the thin base region without recombining and are then swept across the base-collector junction into the collector region. A small percentage don't and these form the base current.
Update to answer a comment:
Once it has been biased isnt Vbe=0.7V. Now when we apply a small ac
signal of the order of milli volt isnt the change negligible?
No, the collector current is exponential in the base-emitter voltage:
\$i_C = I_Se^{(v_{BE}/V_T)}\$
To get a feel for this, consider this question: to double the collector current, how much would \$v_{BE}\$ need to increase?
For example, assuming the bias value is \$V_{BE} = 0.7V \$, increasing this voltage by a mere \$17 mV \$ (an increase of just under 2.5%) will double the collector current.
Another approach:
According to the collector current equation, if we change the base-emitter voltage from its quiescent value by some small amount, the change in collector current is approximately:
\$\Delta i_C = \dfrac{I_C}{V_T} \Delta v_{BE}\$
As a typical example, let the quiescent collector current \$I_C = 1mA \$. At room temperature, \$V_T = 25mV\$.
Then, for these numbers, the collector current changes by 4% when the base-emitter voltage changes by 1mV.
The voltage divider rule between your two resistors does not work like you think because the base emitter junction of the BJT tends to go up to about 0.7V and then not go much higher whilst the current into the base can increase more and more. In other words the BE junction clamps the voltage level between the two resistors to about 0.7V.
When the R1 value is increased to a certain level the voltage at the BJT base lowers down below the 0.6 to 0.7V level and the transistor starts to shut off. At some point the voltage divider will begin to act like normal as the current into the base approaches zero.
ADDITIONAL INFORMATION
Since the OP is not yet quite getting it let me be specific with the examples that were posted. It is correct that at a voltage in range of 0.6 to 0.7V the transistor will begin to turn on.
Let's look at the 20K//1K case in the left picture. Assume for a moment that the transistor base is not connected to the two resistors. By the voltage divider equations the divider voltage is:
Vb = (Vsupply * R6)/(R5 + R6) = (12V * 1K)/(20K + 1K) = 0.571V
This voltage is less than the voltage needed to turn on a transistor so if you would reconnect the transistor base to the divider there will be virtually no current flowing into the base of the transistor and the voltage divider will remain near this 0.571V value.
Next step is to visualize what happens in the above equation when the R5 value is decreased. The divider voltage will increase slowly as the R5 value is decreased.
As R5 decreases more and more the Vb divider voltage will rise up to to the point where the transistor wants to begin turning on. That will be in the 0.6 to 0.7 voltage range. At this point the transistor base begins allowing some of the current from R5 to flow into the base of the transistor.
Be aware that transistors are current mode devices and are actually turned on when the current into the base starts to flow. Below the Vbe threshold the current is nearly zero. As the divider gets past the Vbe threshold the current into the base increases and the transistor starts to turn on.
Ok lets go back and decrease the value of R5 a little more. The lower resistance of R5 allows more current from the 12V supply to flow to R6 and the base of the transistor. The voltage across R5//R6 divider will no longer follow the above equation because the base of the transistor is placing a load on R5 and stealing current so that R6 does not get as much. The nature of the transistor base-emitter junction is that the current into the base can increase more and more whilst the voltage of the base will change only a little.
As I said before the base of the transistor begins to act like a clamp on the voltage divider not allowing the Vb to increase much above the 0.7V level as R5 is made increasingly smaller and smaller. Instead the base current increases to the point that the collector current starts to flow and the transistor eventually turns full on.
The amount of base current needed to turn the transistor full ON will depend on how much collector current is allowed to flow which is limited by components in the collector circuit. The relationship between the base current and the collector current is called the transistor gain or Beta. If the collector current is limited then the transistor will saturate to a Vce of near zero volts when the base current has reached a sufficient level.
It is possible to keep lowering the value of R5 more and more causing the base current to increase more. But beyond the level that caused saturation (Vce near zero) the Vb will only increase slightly and no additional collector current will flow because it has reached the level limited by the components in the collector circuit.
Best Answer
There are many different BJT models, with varying degrees of usefulness in varying circumstances. (See SIDEBAR at bottom.) I'm not going to delve into any of that as it's not necessary in this case. A nice simplification is quite sufficient for your use here.
Ignoring Nth order effects that aren't important here, a BJT's collector current is determined by its base-emitter voltage; here shown using the Shockley diode equation in a highly simplified (active mode) BJT model that ignores the Early Effect:
$$ \begin{align*} I_\text{C}&=I_\text{SAT}\cdot\left(e^{\cfrac{V_\text{BE}}{V_T}}-1\right)\tag{Active Mode}\label{AM} \end{align*} $$
(In the above, \$V_T=\frac{k \:T}{q}\approx 26\:\textrm{mV}\$.)
The important thing to note here is that this equation expresses a relationship between \$V_\text{BE}\$ and \$I_\text{C}\$. A BJT is a voltage-controlled device, as a FET is. (The big difference is that in the BJT case you need to keep supplying additional current into the base in order to deal with recombination. See my explanation here, regarding a PNP, for some details.) This equation works both ways. If you can set \$I_\text{C}\$, then you also know \$V_\text{BE}\$ (see a derivation here):
$$\begin{align*} V_\text{BE}&\approx V_T\cdot \operatorname{ln}\left(\cfrac{I_\text{C}}{I_\text{SAT}}\right) \end{align*}$$
So it should be the case that \$V_o\$ is determined by \$I_\text{C}\approx \frac{V-V_\text{BE}}{R_1}\$. Plugging \$I_\text{C}\$ into the above equation can be solved by the LambertW function as:
$$V_\text{BE}=V-V_T\cdot\operatorname{LambertW}\left(\frac{R_1 I_\text{SAT}}{V_T}\cdot e^{\frac{V}{V_T}}\right)$$
(If you are interested in what the LambertW function is [how it is defined] and in seeing a fully worked example on how to apply it to solving problems like these, then see: Differential and Multistage Amplifiers(BJT).)
However, you could also just make a reasoned assumption about \$V_\text{BE}\$ to start, stick that in to get a refined value and then use that refined value one more time and you'd be close enough for all intents and purposes.
I still haven't mentioned the negative feedback. I didn't want to, at first, because I wanted you to instead focus on the fact that a collector current \$I_\text{C}\$ attempts to force a particular base-emitter voltage \$V_\text{BE}\$, just as a particular \$V_\text{BE}\$ would attempt to force a particular \$I_\text{C}\$.
In the current mirror, you drive a collector current into one BJT, which then forces a particular \$V_\text{BE}\$ for it, which is then passed along as a voltage to a second BJT as a \$V_\text{BE}\$ signal for it, which then attempts to cause a particular collector current for this second BJT's collector load.
So let's get down to the business of the negative feedback, now. Let's start with the schematic:
simulate this circuit – Schematic created using CircuitLab
The equation is pretty simple:
$$\begin{align*} V_\text{BE}&= V-R_1\cdot I_{R_1}\\ &=V-R_1\cdot I_E\\\\ &= V-\frac{\beta+1}{\beta}\cdot R_1\cdot I_\text{C}\tag{Eq. 1}\label{eq1}\\\\ \textrm{or,}\\\\ I_\text{C}&=\frac{\beta}{\beta+1}\cdot\left(\frac{V}{R_1}-\frac{V_\text{BE}}{R_1}\right)\tag{Eq. 2}\label{eq2}\end{align*}$$
In equation \$\ref{eq1}\$ if \$I_\text{C}\$ increases, then clearly \$V_\text{BE}\$ decreases. But from the \$\ref{AM}\$ equation discussed earlier, a decreasing \$V_\text{BE}\$ must imply a lower \$I_\text{C}\$. The contrary relationship between these equations amounts to negative feedback. In equation \$\ref{eq2}\$, relating things the other way, you can see that if \$V_\text{BE}\$ increases then the collector current decreases as a result. But once again from the \$\ref{AM}\$ equation a lower collector current implies a smaller \$V_\text{BE}\$. So again, negative feedback.
In effect, \$R_1\$ (as arranged here) is providing negative feedback.
Now return back to the BJT current mirror case. If you ignore the fact that there are some errors caused by having to supply base currents, some errors caused by mismatched \$\beta\$ values in the two BJTs, some errors possibly caused by different temperatures in the two BJTs, some errors also caused by different saturation currents (and therefore different \$V_\text{BE}\$ given the same collector currents), and some further errors that may occur because of the Early Effect due to different \$V_\text{CE}\$ values between them...
If you can get past all that then sinking a current into one BJT leads to a necessary \$V_\text{BE}\$ for that BJT, which drives the other BJT's base-emitter voltage causing a corresponding collector current in it. The second BJT mirrors the collector current in the first BJT!
Similar arguments apply to the MOSFET case, too, though the equations and unaccounted errors and thermal behaviors are different.
SIDEBAR: These models include Ebers-Moll (three 1st-level DC models that are all equivalent and described in my answer elsewhere on EE.SE, plus follow-on modifications to Ebers-Moll in order to handle everything from lead resistance to base width modulation effects, to Gummel-Poon (provides a unified view of the Early and Late Effects and more, as well) and further modifications, then to VBIC, and still later ones which are used by FAB personnel for IC designs. All useful BJT models are based upon some kind of understanding of the the physics involved. But even that physics, itself, is yet another set of simplifications about a deeper reality. For example, even the most thorough attempts will incorporate an assumed gas cloud model of conduction band electrons, ideas of mean free path, and so on. The reality is vastly more complex than that. But once you move fully in that direction, you leave behind electronics entirely and have moved into profound physics.