Why don't I ever have to calculate how much voltage is left after the drop?
Maybe you haven't yet, but you will at some point. You'll need to be able to use that to figure out other voltages in the circuit. The way we were taught was around Kirchoff's Voltage Law, which basically said that the sum of voltages in any "loop" including the source would be 0. I've been working on analyzing a BJT transistor, and knowing the voltage 'left over' is helpful (at least in my circuit) for figuring out the Base-Emitter voltage and the Collector-Emitter voltage that are "left over" after resistors' voltage drop.
So, if you have more than one element in the circuit in series, the voltage "left over" for the 2nd (or 3rd or 4th) element in series can be useful to know.
In your friend's example, with 3 identical resistors in series, they will each drop 1/3 of the total source voltage. Because they drop, in total, the full 10V across all of them. So Resistor 1 has 10V "left over" since it is hooked up to the batt. It has a 3.33V voltage drop. Resistor 2 has 10V-3.33V - 6.66V "left over". It also drops 3.33V. So Resistor 3 has 6.66V - 3.33V = 3.33V "left over". It drops the remaining 3.33V so that there is 0V "left over".
The energy potential from the battery is always all used up by the stuff connected across the + terminal to the - terminal of the battery. The fact that you connect stuff to the - terminal means that there will be zero energy at that point.
You can only measure voltage across two points... one of those points might be ground, which is sort of the 'voltage left over' concept. Or you can measure across two points to measure the 'drop'
Michael