How does a 'voltage drop' work?
This is because of what you might call 'differential impedance': the change in current per unit change in voltage. A resistor's differential impedance is linear: for every extra volt across it, a fixed number of additional amps flow through it, determined by that resistor's resistance value.

Semiconductors don't exhibit linear differential impedance; a unit change in voltage won't always create the same change in current. If you plot voltage vs current for a diode, you'll see something like this:



This is what creates the diode's characteristic voltage drop. The curve is sharp enough that it can mostly be approximated by saying it has a constant voltage drop, but you can see that that's not true - the voltage drop will gradually increase with current. This is easier  to see on a log scale graph:



The upshot of all of this is that in a circuit with, say, a resistor and diode in series, they will find an equilibrium, where the current and voltage through each part forms a point on their respective IV curves. Since a diode's IV curve is very steep, the voltage across it shifts very little with current, and most of the voltage will be dropped across the resistor instead.