Reverse recovery time matters because it's like the time it takes for the diode to switch from blocking to conducting. Think about it, when the inductor is being powered normally, the diode must be blocking. But the instant power is cut and the inductor starts trying to dump energy, you want the diode to begin conducting as fast as possible. THe longer it takes to conduct, the higher the voltage spike will be and the more likely something is going to fry (remember, the voltage will instantly spike to a level to force current to flow, aka where something has got to give). Until the diode does turn on and allows low voltage, high current energy from the inductor to dump, a high voltage, low current energy is flowing through something else. Who knows what it might be? But it's probably not something that's supposed to handle it.
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Voltage drop is important because it helps define what level the BEMF is clamped to. In your scenario with a diode in parallel with the inductor, once the diode conducts the BEMF Is clamped to the forward voltage drop. So there's not too much of a difference between the clamping voltage 0.7V of a regular diode and 0.3V schottky.
For an H-bridge though, it matters a bit more. Obviously you can't put a diode in parallel with a motor in an h-bridge because current can flow both ways. So you either have a short in one direction if you try and clamp the BEMF when cutting power in one direction only, or a short in both directions if you try and clamp the BEMF when cutting power for both directions. So instead diodes are placed in parallel with the switches/transistors. This provides the same flyback current path as before (from one the terminal of the motor to the other), but now it needs to take the path through two diodes instead of just one and the result is you don't have a shorts that would affect normal operation. There is a catch though...instead of clamping the BEMF to the forward voltage like your scenario, it now clamps it to Vsource+Vforward. It is only capable of clamping the BEMF to a voltage ABOVE the source/battery voltage. Do you see how a high forward voltage drop coudl be a problem in that case?