Confused about higher potential at junction in real(practical) diode
The built-in potential will block the current from flowing. If the N-side is at a higher voltage, then the current trying to flow from P to N will be stopped by the positive voltage barrier. Ie. positive charges going from P to N will not be able to, because they are driven away by the positive N-side (the other way with negative charges happens as well).

If you apply a more positive voltage to the P-side compared to the N-side, then the N-side will be less positive than before, and it is slightly easier for the positive charges to pass over this positive barrier, so the current increases.



The voltage source in the model you presented however has a different meaning. As the diode has an exponential characteristic, it is a pain to work with analytically. So they approximate the IV-curve with a model. But this is not necessarily a representation of the built-in potential, although it can (but not necessarily) be linked to it. The voltage source there is purely to approximate the black curve with the corresponding red curve.