I'll try to offer an explanation as short as possible why a short run of smaller wire is acceptable coming out of a device, but not for long runs.
First, to understand resistance and the American Wire Gauge (AWG) standard, where 10-ga. wire is sized such that 1000 feet equals 1 ohm of resistance. For each 3 gauge sizes smaller or larger, the resistance doubles or halves. Therefore, 7 gauge is .5 ohms per thousand feet, and 4 gauge is half of that (0.25 ohms per thousand feet). This doesn't sound like much resistance, and I certainly don't know how many amps a windlass draws (hopefully someday I'll need to know), but the 8-ga. wire supplied by the winch will safely handle 40 amps continually. To determine how much voltage-drop will be wasted across the wire, first understand that current is the same in the entire loop to the winch. That means if 40 amps is flowing through the breaker, there will also be 40 amps flowing through the positive, negative wires and switch. If the entire run (positive and negative) wires total 50 feet for example, then the resistance in 50' of that .25 ohm per thousand will be .0125 ohms. (50'/1000'=.05) (.05x.25 ohms per thousand=.0125 total ohms). When you multiply .0125 ohms by 40 amps, you get 0.5 volts. This means that your windlass does not receive 12 volts, but 11.5V which will operate better than voltage somewhere in the 10 volt range if 8 ga. were used. These voltage drops will be even more severe if greater than 50' is used, or if the motor draws more than the 40 amps I estimated. This is why a motor can get away with being supplied and even operated with a couple of feet of smaller wire. The voltage drop deducted from the 12 volts (wasted across the wire) is very negligible for short runs, but not 50 feet. I hope I explained this well enough without involving too much theory.