Theo Buehler did a great job relating Types A, B, C, and E. Going off of Stefan Witzel's new answer, I'd like to point out that in Munkres's Topology in section 81 (page 505 of that link), he defines an action to be properly discontinuous if for all $x\in X$ there is a nbhd $U$ s.t. $g(U)\cap U = \emptyset$ unless $g=1$. So if we tweak Type D from the original question to conclude $g=1$ rather than just that $g$ fixes $x$, then we recover Munkres' definition and it's no longer as trivial as Theo's answer showed the old Type D definition was. I think Munkres' definition is the one which I think point-set topologists would use. It's nice because you don't need to assume a topology on $G$, but of course you could just put the discrete topology on it. Perhaps the other definitions are more popular in the literature of Riemann surfaces, and the difference may be because of standing hypotheses in that field, since they often care most about the case of Fuchsian groups. Certainly Munkres' definition implies Type E and Type C
Munkres also points out that the quotient map $\pi: X\rightarrow X/G$ is a covering map iff the action of $G$ is properly discontinuous.
An exercise in section 81 gives: Let $X$ be locally compact Hausdorff and let $G$ act freely (i.e. fixed-point-free). Suppose that for each compact $C \subset X$ there are only finitely many $g\in G$ s.t. $C\cap g(C) \neq \emptyset$. Then the action of $G$ is properly discontinuous and $X/G$ is locally compact Hausdorff. So this tells you when Type B implies Munkres' definition.
Now let's relate Munkres' definition to Type A and Theo's answer. Using Theo's various propositions and corollaries it's not hard to see that if $X$ is locally compact Hausdorff space and $G$ is any group (which we'll equip with the discrete topology) then Munkres' definition implies Type A. Conversely, if $X$ is locally compact then a proper action of a discrete group must be of Type B (by Theo's comment) and this implies Munkres' definition because local compactness lets us get from $g(K)\cap K = \emptyset$ to $g(U)\cap U = \emptyset$.