Let $G$ be a group and $S$ a set with an action of $G$. I'll assume it's a left action, so that $g(h(s)) = (gh)(s)$. If $O_s \subset S$ is the orbit of $s \in S$ (that is, $O_s = { g(s): g \in G }$), and $G_s \subset G$ is the stabilizer of $s$ (so $G_s = \{g \in G: g(s) = s \}$, a subgroup of $G$), then there's an isomorphisms of sets-with-$G$-action
$G/G_s \to O_s$
given by $gG_s \mapsto g(s)$. You can check that this map is well-defined: if $gG_s = hG_s$, then $g(s) = h(s)$.
As you noted, $G_s$ is not a normal subgroup of $G$ (yes, $gG_s g^{-1} = G_{g(s)}$), so what I mean by $G/G_s$ here is the space of left cosets $\{g G_s: g \in G\}$ of $G_s$ in $G$.
This coset space has a left action of $G$ by translation (that is, $g(hG_s) = gh G_s$), as does the set $O_s$ (the action is inherited from the action on $S$) and the map above is $G$-equivariant (starting with a left coset, you can act by $g$ and then apply the map or apply the map and then act by $g$, and get the same element of $O_s$ either way).
The First Isomorphism Theorem is a statement about a map of groups $f: G \to H$, not about an action of a group on a set, so a priori is dealing with a different scenario. You can probably find ways to see one of these reducing to the other in special cases, but no particularly illuminating setups come to mind at the moment...