Why does a voltmeter show voltage drop

There is one thing that has been missed out from your 'classic resistor example' - the internal resistance of the battery.

If you simply measure the battery voltage with a voltmeter you get a higher reading due to the fact that there is no (or very little) voltage dropped across the internal battery resistance.

The voltage measured across a load resistor (or bulb) OR the terminals of the battery will be smaller by an amount equal to I * R(int).

The connecting wire has very little resistance and so does not contribute to a significant voltage drop between the terminals of the battery and the resistor so the voltages at these two places are, in practice, the same.

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.