voltage drop calculations single phase ac

In voltage drop calculations, in one of the methods in case of single phase ac, the formula is given as follows As you can see, R is resistance per unit length multiplied by length by current and then by 2. Why the 2? Now the reasoning they give us for this, is that the current returns through the neutral, so the length of the path is of both the live and neutral wire.

This makes zero sense to me. To my understanding, and according to this answer to a different question on here :

https://electronics.stackexchange.com/a/38694/218615

To quote : "The voltage is a difference between the electric potentials of two conductors. Hence, to change voltage, only one of the potentials has to change (although both can). In AC power only one of the wires (live/phase) changes it's potential, while the potential of the other one (neutral) remains constant."......"in properly constructed power network the neutral wire is maintained at a potential level close to ground potential, there is nearly no voltage between the neutral and the ground. Hence, touching neutral will not cause current to flow through human body into ground." End quote.

So according to this quote, and what I understand, the voltage drop would be the difference in potential between point A on the wire and point B relative to the neutral wire which is almost always zero or the reference. Here's an illustration: Say this is a 220 volts ac single phase source. Voltage at A is say 220, which means Voltage at A compared to neutral is 220. And when current flows from A to B , a voltage drop occurs due to wire resistance so voltage at B relative to neutral is say 218 and not 220. And so the voltage drop would be 2 volts. Is my understanding incorrect? What does the neutral have to do with the voltage drop? Why do I multiply by 2? I feel there is a basic pivotal thing I am missing here. Any help is appreciated. Thx 