What happens if a voltage regulator is driven with input voltage a lot higher than nominal output?
Will it then just short 7 V to ground in order to maintain the 5 V under load?

It doesn't short anything to ground, but besides that you got it right. If the input is 12V and the output is 5V then you'll have 7V across the regulator. Since the load current goes through the regulator, it will dissipate according to P=IV. So if you got 1.5A and 7V across the regulator, then indeed it will dissipate 10.5W and require a heat sink.
For better efficiency, at this kind of high current it is much more convenient to use a switching regulator. For example you can use these ready-made ones which are very simple to apply, just read the datasheet and add the required input/output caps. If you wonder what's inside, these show the internal schematic in the datasheet. There's also a search engine.
At the rated efficiency of 90%, losses are 10% so with an output power of 5V*1.5A=7.5W the switching regulator will dissipate only 0.75W, which means you don't need a heat sink. It is more expensive than a 7805 but when you compare to the cost of a 7805 plus a large heat sink, it's a much better solution. Also it will draw much less current from the input.