What happens when Schottky diode is protecting a pin from overvoltage?
where the diode is connected (in the appropriate direction) between the pin and the +5V Vcc, doesn't that mean that you're dumping current into Vcc when (for instance) +7V is being applied to the pin?


Yes, you dump the energy into VCC potential, raising its voltage.

In ESD (electrostatic discharge) situations this is OK, because there is usually at least an 100nF cap present on VCC/GND (and often >=10µF in parallel). ESD discharges are modeled by a small but high voltage capacitor - much smaller than 100nF, so the voltage rise is very small, too.

The other case is a (permanent) connection to a higher voltage. That can burn out the diode, the battery/voltage source and the electronics it was supposed to protect, by raising VCC above damage level. 

This is where additional protection like series resistors and TVS diodes come into play.