Stranded copper has different resistances based on the strand count and gauge,
see here, it's around 2.6 to 2.7Ω/1000' for 14AWG.
Lets say you have a 6' jumper that uses a strand providing 2.7Ω/1000', 6 foot has a resistance of 0.0162Ω. Now double that, because you have two conductors, so 0.0324Ω.
Ohms law tells us that 20A of current through 0.0324Ω will drop a voltage of 0.648v. Ohms law also tells us that with 20A at 0.648v, 12.96w of heat is generated.
On a 13.8v supply, losing over half a volt to the cable means your radio only sees 13.32v, and although that may be acceptable, the cable insulation needs to dissipate that 12.96w of power, which it probably can as it is about 90mW per inch of wire.
Whether the cable can stay cool or not depends on the duty cycle of your transmissions and on the area, thickness and thermal conductivity of the insulation used on the wire. If the insulation can only dissipate so many watt/seconds and you are giving it more that that, it gets warmer and warmer until either the larger temp difference supports amount of energy transfer and the temp stabilizes, or the wire coating melts off. I personally would want a bigger wire if running 20A. There is no reason to give the wire almost 5% of your energy.