I believe I understand how an excessively high impedance can damage a final output power MOSFET (voltage rises until avalanche breakdown). But how does an excessively low impedance do that? If the impedance is too low, won't the current simply go through the load, bypassing damage to the MOSFET?
Step away from RF for the moment, and think about the basic operation of a single-supply transistor amplifier, or even a voltage regulator.
Ignoring the details of how the gate is controlled by the input signal, the current flows in a single simple circuit: through the supply, the transistor, and the load. Therefore, however much current flows through the load, also flows through the transistor. If the load has a lower impedance than intended, then more current flows through the transistor as well as the load than intended, and therefore the transistor gets hotter. The above circuit is a DC circuit, but exactly the same principle applies to AC/RF circuits — scale up instantaneous currents and you scale up waste heat.
Thus, the transistor is still performing its function — no electronic failure has immediately occurred as a result of the load impedance — but the additional heat may raise its temperature to the failure point (depending, of course, on the transistor's limits and the surrounding heat-sink / cooling system).