Your definition implies that $$ \tilde S = \bigcup_{p\in M} TM_p^{G_p}. $$ In particular, $\pi(\tilde S) = M$, and $\tilde S$ will be a submanifold of $TM$ iff the dimension of $TM_p^{G_p}$ is the same for all $p\in M$.
$G$ being connected won't necessarily make this happen: Another counterexample is $S^1$ acting by rotation on $S^2$, fixing the north and south poles, $n$ and $s$. $\tilde S$ is then $T(S^2\setminus\{n,s\})\cup \{n,s\}$, and the points $n$ and $s$ do not have neighborhoods homeomorphic to open balls.