Falling and reflected impulse waveform - short circuit and void
1) In the open case, it comes to to conservation of charge.  That is, since it's a open circuit, no current can flow, but you have an incident wave impinging on the open circuit.  Therefor there needs to be an equal current flow out of the open circuit back towards the source.

2) In the short circuit case, it comes down to the fact that the voltage has to be zero at the destination (since it's a short circuit).  And so you get a negative going wave (assuming the incoming wave was positive going) flowing out of the short back to the source.

Wikipedia has a good explanation, here: 
https://en.wikipedia.org/wiki/Reflections_of_signals_on_conducting_lines#Open_circuit_line