Voltage drop of a non-ideal voltage source
The current I2 should be equal to I2 = U02/R2 = (E2-V2)/R2, but I do not understand why this would be the case.


This is not correct. 

The current through the resistor is determined by the voltage across the resistor. Since the resistor is not connected to node 0, then the voltage at node 0 has no role in determining the current through the resistor.

If I understand your notation, you should have

$$I_2 = \frac{V_A-V_2}{R_2} = \frac{U_{A2}}{R_2}$$