My, that's a lot of questions.
Presumably the diode model you instructor wants you to use is an ideal diode with 0.7V drop in series with a 4 ohm resistor. So, when the diode is conducting, it behaves like a voltage source with a resistor in series. When the forward bias is less than 0.7V it does not conduct.
Is the 4 ohms per diode negligible in comparison to 1.5K? Well, it's more than 0.5% for two diodes so it will drop tens of mV. That might be negligible or it might not be, depending on the application. Since the instructor gives you the value, I suggest it might not be negligible in terms of getting the correct answer.
Two such diodes in series behave like one 1.4V diode with 8 ohms in series.
why does the power supply only "consider" the 1.6V drop across the LED and sends current accordingly
Note that power supplies don't "send current," instead they send voltage. The load resistor then "draws current" based on Ohm's law (or for diodes, based on the V-I curve.)
I think your confusion is caused by the concept "nonlinear resistance." Diodes don't actually turn on and off, instead they have nonlinear voltage/current behavior. Diodes don't behave as resistors, instead their current is determined by the applied voltage, and described by (oh no!) an exponential function. Because of the LED nonlinear resistance, even a simple LED with series-resistor isn't perfectly easy to understand.
Your circuit will be doubly-confusing because you're fighting two "nonlinear resistors" against each other: the LED's nonlinear curve, versus the nonlinear curve for the whole diode-chain. Nasty!
:)
Here's one way to look at it. Suppose we slow things down by adding a large capacitor from NODE1 to GND, like 3,300uF. Next, when we suddenly connect the battery, the voltage on the capacitor starts rising. The capacitor voltage is also across the LED and the diodes. Eventually the voltage will arrive at the "fast rising" part of one of the diode graphs. In this case the LED arrives first (it's around 1.0V for red-color LEDs, higher for other colors.) The fast-rise part of the diode chain's voltage is around .4V for each diode, times nine, so roughly 3.6V, much larger than the LED volts. As the capacitor voltage rises, the LED "wins." The rising voltage will level out as soon as the resistor's Ohm's law behavior gives the same current as the V-I equation for the LED.
In other words, the diode-chain cannot draw significant current until your LED voltage goes above 3.6V!! This won't happen with a red LED and a 2.7K resistor.
However, if you'd used a white LED and a 100-ohm resistor, the diode-chain WILL draw significant current. If a white LED draws 30, 40, 50mA, the voltage can climb well above the usual 3V seen on white LEDs.
So, the answer to your question is different for different color LEDs!
See? Nasty.
In cases like these, the only way to make completely accurate predictions is unfortunately to abandon simplified mental models. Instead, write and solve equations. (This one has two exponential equations, one for the LED and another for the diode-chain.) Or, use a circuit simulator or Spice program which is invisibly solving equations for you in the background. Adding a capacitor and imagining slowly-changing conditions can take you far in grasping nonlinear electronics. But sometimes it's not obvious where that capacitor should be placed, or which nonlinear component will dominate.
Best Answer
Consider this diode (about 0.7V Vf) in parallel with an LED (about 2V Vf). Of course the actual voltage across each will be the same.
Vf, as we are talking about it here, is not the real forward voltage except in very particular circumstances, it's an approximate voltage that you would measure when some particular (sensible in the context) current flows through the diode. So it's a characteristic of the part.
simulate this circuit – Schematic created using CircuitLab
When the diode is in parallel it "hogs" almost all (99.9%) of the 10mA current, and the forward voltage of both ends up being almost exactly the same as the diode alone.
The LED still conducts a bit of current according to the simulation, about 10uA, but without higher forward bias it cannot conduct much current .
A simplified way to look at it would be that the diode with the lower Vf takes all the current and the one with the higher Vf takes none of the current AND the actual voltage equals the smaller Vf.
That may be accurate enough, however that gets progressively less accurate the closer the two Vfs are to each other. For example if we simulated 2 1N4148s in series (about 1.38V Vf) in parallel with the same LED only 94% of the current would go through the diodes, and 6% through the LED. If the two diodes were identical, the current would obviously (by symmetry) split equally, and the Vf would be about the same as one diode (in reality it would be a bit less).