Why can you short voltage sources and open current sources and not vice versa?

This method of finding $R_{\rm Th}$ follows from the linearity of all the Kirchoff's Law equations (from which Thevenin's theorem is derived.) The value is $R_{\rm Th}=V_{AB}({\rm open})/I_{AB}({\rm short})$. If you multiply all of the equations by a constant, you get the same answer when calculating $V_{AB}$ and $I_{AB}$ by Kirchoff's Laws. All of the terms in the circuit equations with current in them are proportional to $I$ and the voltage source terms are proportional to $\varepsilon$. If you multiply all equations by an arbitrarily small constant (still getting the same answer), that's like setting the EMF's to zero, which is like shorting them out (no voltage drop across them). At the same time it's also like making all the currents approach zero, which is equivalent to removing the current sources from the circuit as well. So you get the same equivalent resistance when you short the voltage sources and open the current sources.

In contrast, if you opened the EMF's, that would give you a fundamentally different circuit- there could then be any potential difference across a voltage source's terminals, not an arbitrarily small one. Similarly, shorting the current source changes the nature of the network.