Simple voltage drop question
Your answers: 


Yes, the current flows through both lengths of cable so you have to account for both.
Only voltage drops.  Current is drawn through.
Basically I wouldn't go longer than a drop that would take it below the charge threshold of the batteries / charge circuit.


Basic ohms law is your friend:

For 1 amp of draw:


14AWG has a resistance of 2.525mΩ/ft
So 200' has a resistance of 0.002525 * 200 = 0.505Ω
V=IR
Therefore V=1 × 0.505
Which equals 0.505V


For the full 5 amps of draw:


V=5 × 0.505
= 2.525V


Which is obviously quite a drop.

This table has a nice list of the resistances of different cables.  As you can see increasing the cross-sectional area (decreasing the AWG) decreases the resistance.  Even just going from 14 to 12 AWG has a big difference (1.588mΩ/ft as opposed to 2.525mΩ/ft), and fatter cables are even better.