Not understanding NPN transistor in circuit
I think the potentiometer counts as part of the voltage divider

You can use a potentiometer as a voltage divider but here it is not used that way. Here the voltage across the potentiometer is somewhat constant (9 V (battery) - 0.7 V (Vbe of the NPN) - 2 V (LED) = 7.3 V.

Yes I am ignoring the 10 kohm resistor, I pretend that the potentiometer can vary between 10 k ohm and 1.01 Mohm.

As that 7.3 V across the potentiometer is somewhat constant, the current through the potentiometer will vary with the resistance. At 10 kohm we get 7.3 V / 10 k ohm = 730 uA, at 1.01 Mohm we get 7.3 uA.

So the potentiometer in this circuit behaves like a "variable current source". That current flows into the base of the NPN. The NPN transistor will then amplify this current by a factor 100 to 500 depending on the NPN. That will then vary the intensity of the LED on the right.