Two well-known definition for the real $n$-dimensional projective space $\Bbb{RP}^n$ (as a smooth manifold) are as follows:
Taking the quotient of $\Bbb{R}^{n+1}\setminus 0$ by the equivalence relation $x\sim \lambda x$ for all $\lambda \in \Bbb{R}^\times$, which is the same as taking the quotient in the group action of $\Bbb{R^\times}$.
Taking the quotient $S^n/{\pm}$, that is, gluing antipodal points in the $n$-sphere.
It is not very complicated to show the equivalence between the two definitions. However, one may think of the subgroup action of $\Bbb{R}^\times_+$ (positive reals) on $\Bbb{R}^{n+1}\setminus 0$, and taking the quotient of something similar to
$$\frac{(\Bbb{R}^{n+1}\setminus 0)/\Bbb{R}^\times_+}{\Bbb{R}^\times/\Bbb{R}^\times_+} \cong \frac{\Bbb{R}^{n+1} \setminus 0}{\Bbb{R}^\times }, $$
which form is similar to the third isomorphism theorem. Is there a simple way to formalize this isomorphism? (As topological groups, say)
Here's one way to think about this. Let $G$ be a group acting on a set $X$, and let $N$ be a normal subgroup of $G$. You then have an action of the group $G/N$ on the quotient set $X/N$, and it satisfies $$X/G = (X/N)/(G/N).$$ The proof of this is straightforward, and I guess can be thought of as a version of one of the isomorphism theorems for group actions.
In your situation, $G$ is a topological group acting continuously on a topological space $X$, and the normal subgroup $N$ is a closed normal subgroup. Since $N$ is closed, the quotient group $G/N$ is a topological group, and everything above goes through in the topological category. But I think it is clearer to think of it as being a fact about group actions on sets.