I understand how an H Bridge works, but the common ones use pnp transistors on top and npn on bottom. In L298N's datasheet it is attached this circuit and I just can't figure out how the Vs is going to supply the motor if there is a npn on top. I know that the emitter voltage of the transistor on top is depending on base voltage as follows: Ve = Vb - Vbe, so where is the Vs? Also, shouldn't the base voltage (MCU output voltage) be greater than the collector voltage (Vs) in such a circuit?
The high side NPN is used as an emitter follower.
Yes, it cannot provide up to VCC, but about a Vbe drop less.
The MCU interface does not affect the output, because there is an internal voltage translation to VCC, so that is why the motor output is VCC minus the Vbe drop.
Have a look in the datasheet of the L298, at the bottom of page 3. There the source saturation Voltage is mentioned. This is the voltage that is dropped inside the L298 when an output is high and a current is flowing:
Ideally we don't want any voltage to be dropped, we want 0 V or more realisticly: as little as possible.
You're correct that because it is using an NPN there, you would need a higher voltage than +Vs to be able to make those NPNs lift the output voltage up to +Vs. But the L298 isn't designed that way so it simply cannot pull its output all the way up to +Vs. In the table we can see that typically at 1 A flowing, the L298's outputs drop 1.35 V.
Also note that things aren't much better on the "low" side either, the Sink saturation voltage is 1.2 V at 1 A.
If you want lower voltage drops than this, I suggest using a more modern IC. This L298 is ancient, I am surprised that ST still sells them!