The built-in potential, $V_{ic}$, is obtained by analyzing the equilibrium situation, where drift and diffusion currents cancel each other out in the depletion region. This built-in potential is necessary for current not to flow in the system.
When you apply an external voltage you expect some current to flow, so the built-in potential must be altered, such that a net current can flow in the system. The question is, why is the change in the built-in potential exactly that of the applied voltage?
The reason is that nowhere else in the system can you have such a potential drop. The non-depleted regions of the p and n sides have such large carrier concentrations that a very small (negligible) electric field on them is enough to drive a current across the system. The negligible electric field results in a negligible potential drop across the non-depleted p and n regions.
Put another way, the depletion region has much less carrier concentration, and therefore presents the largest resistance to current flow in the system. The electric field associated with the voltage $V$ adds to the built-in field in the depletion region, such that the proper amount of current can flow.
Of course, the better way to approach this problem is to start out with the equations that govern transport in semiconductors (drift, diffusion, generation+recombination, etc.) and treat the applied voltage as a difference between quasi-Fermi levels on the two ends of the junction.
The numerical solution to the set of differential equations will show that the results of the depletion model (for the depletion region width) are a good approximation if the built-in potential $V_{ic}$ is replaced by
$$ V_{ic} - V $$
In introductory textbooks, this result is simply quoted as if it was obvious. It turns out is not that obvious, but it is a pretty good approximation for this situation.
