Do you know what an old-fashioned fuse consisted of? Just a bit of thin wire.
Can you recall from your schooling the formula for Volts, Watts and Amperage? The current I in amps (A) is equal to the power P in watts (W), divided by the voltage V in volts (V)
If you run a current on too thin a cable, the cable offers impedance (resistance) to the current, causing some of the current to turn into heat, thus diminishing the effective current delivered to the endpoint.
It's up to you if any of that matters to you or if you think the manufacturer was just blowing smoke up your keester to specify requirements.