Fundamental group of quotient spaces of $SO(3)$

This is cool. Are you doing chemistry?

So, a slightly more standard notation might be $G_1\backslash SO(3)/G_2$ -- this indicates that you're left-multiplying by elements of $G_1$ and right-multiplying by elements of $G_2$.

A "properly discontinuous" action of a group $G$ on a space $X$ is one where every point $x\in X$ has a neighborhood $U$ such that $g(U)\cap U=\emptyset$ unless $g=1$; that is, $G$ not only acts freely (no nonidentity element has fixed points), but the nonidentity elements of $G$ take every point "sufficiently far away from itself". This may seem vacuous, but in fact when you're working with Kleinian groups and things this can be an important condition. However, in your case what is immediately true (since your space $SO(3)$ is Hausdorff and your groups are finite) is that any free action is automatically properly discontinuous. So the only thing you need to check is that your actions have no fixed points.

So I guess the question becomes, do you know how to obtain this group $\Gamma$ explicitly? If it's cyclic, it probably will be taken to be a subgroup of the circle, which acts freely on $S^3$. (This is $S^1\subseteq \mathbb{C}$ acting on $S^3\subseteq \mathbb{C}^2$ by multiplication.)

If $\Gamma$ doesn't act properly discontinuously, things are harder.