Electronic – Explain in layman’s terms Vgs and Vgs(th) of MOSFET’s

Point 1 and point 2 are quite accurate I would say. For point 3, I always search for the following graph when I'm looking for the likely conduction capabilities of FETs: -

It tells me that if I put 3.5V on the gate and I want to pass 10A drain current there is likely to be 0.5V dropped across the device leading to a power dissipation of 5W. This is an on-resistance of 50 m\$\Omega\$. However, if i used a gate drive of 10V, I can expect the voltage dropped by the device to be about 0.25V i.e. a power dissipation of 2.5W or an on-resistance of 25 m\$\Omega\$.

Regarding the state of "fully-on" you can see that there is no magic state that is reached at some arbitrary gate voltage but I'd say that at about 4.5V gate drive, the benefits of driving with a higher gate voltage are diminishing.

Figure 3 in the data sheet (which shows the graph which you deduced the on resistance variation of 10 m\$\Omega\$) is a derivation of the graph (figure 1) in my answer. Obviously, Figure 1 carries more information because it covers a wider range of gate voltages.

Looking into the drain, the small-signal resistance is $$r_{id} = r_o = \frac{\lambda^{-1}+V_{DS}}{I_D}$$ if the source is at AC common (common-source configuration).

If the AC resistance from source to common is \$R_{ts} \ne 0\$, the small-signal resistance looking into the drain is

$$r_{id} = r_o \left(1 + \frac{R_{ts}}{r_s} \right) + R_{ts}$$

where

$$r_s = \frac{1}{g_m}$$

Looking into the source, the small-signal resistance is

$$r_{is} = r_s$$

The above assumes the body is connected to the source.

I understand why r02 is in parallel with Rf, but what is the 1/gm2 resistor doing there?

The lower right circuit is drawn oddly and further, seems to mix AC and DC sources which is an error.

If I were teaching this circuit, I would draw the AC circuit, with Q1 and Q2 replaced by their small-signal T-models, as follows

^{simulate this circuit – Schematic created using CircuitLab}

Now, is it clear why \$r_{s2} = \frac{1}{g_{m2}}\$ is there?

Edit: on second thought, I don't understand also why the first r0 is at the drain of the first transistor, and the second r0 is at the source of the second transistor.

\$r_o\$ connects to the drain and source.

Since, for Q1, the source is grounded, \$r_{o1}\$ connects from D1 to ground.

Since, for Q2, the drain is AC grounded, \$r_{o2}\$ connects from S2 to AC ground.