I think I found the answer, so I'll post it here.
Firstly I'd like to prove that the map $\phi :H \rightarrow\pi(H)$ is bijective, where $\pi$ is the map $G \rightarrow G/N$. This $\phi$ bijectively maps subgroups of $G$ containing $N$ to subgroups of $G/N$.
Injective: $\phi(H_1) = \phi(H_2) \implies \pi(H_1) = \pi(H_2) \implies \pi^{-1}\pi(H_1) = \pi^{-1}\pi(H_2)$. Now by applying $2$ to each side of this equality, I have that $H_1 = H_2$.
Surjective: Take $A/N \leq G/N$. By $3$ I have that $\pi\pi^{-1}(A/N) = A/N$.
Furthermore, $A/N \leq G/N \implies N \triangleleft A \leq G$. Therefore $\pi^{-1}(A/N) = A$ which is a subgroup of $G$ containing $N$. so $\phi(A) = \pi(A) = A/N$. Therefore the map is also surjective.
Finally, I want to show that normal subgroups in this map "match" with each other. This immediately follows from $1$.
In general, if $N\leq K\leq G$, and $N\triangleleft G$, then $N\triangleleft K$: to see this, note that for every $g\in G$ we have $gNg^{-1}=N$, and therefore for every $k\in K$ we also have $kNk^{-1}=N$ (since $k\in G$ as well). So the fact that $N\subseteq HN\subseteq G$ and that $N\triangleleft G$ guarantees that we also have $N\triangleleft HN$.
On the other hand, because you do not know if $N\subseteq H$, then you cannot state that $N\triangleleft H$: in order to be a normal subgroup, you must be a subgroup; and in order to be a subgroup, you must be a subset. Since we do not have any information about whether $N$ is contained in $H$ or not, you cannot assert that $N\triangleleft H$; in particular, "$H/N$" may not even make sense.
(Note however, that if $N\subseteq H$, then you will have $HN=H$)
I do not understand what you mean when you say "... every normal subgroup of $H$ belongs to $\mathrm{ker}(\phi)$." What does it mean for a normal subgroup to "belong" to something? It is not true in general that every normal subgroup of $H$ is contained in $\mathrm{ker}(\phi)$; if you've somehow reached that conclusion, then your argument is incorrect.
My view is that the "right" way to think about the Second Isomorphism Theorem is as a counterpart to the Lattice (or Fourth) Isomorphism Theorem. The Lattice isomorphism theorem tells you that if $N\triangleleft G$, then there is a one-to-one, inclusion preserving correspondence between subgroups of $G$ that contains $N$, and subgroups of $G/N$; and moreover that this correspondence identifies normal subgroups with normal subgroups. And that this correspondence is induced by $\phi$; that is, it also tells you what $\phi$ does to subgroups of $G$ that contain $N$.
This should lead one to wonder: "Okay, that's what is going on with subgroups of $G$ that contain $N$. I understand what $\phi$ does to subgroups of $G$ that contain $N$. What about other subgroups of $G$? What does $\phi$ do to them?"
And the Second Isomorphism tells you: what happens to $H$ is the same thing as what happens to $HN$, which happens to be a subgroup of $G$ to contains $N$; namely, $H$ is mapped to $H/(H\cap N)$, and this is isomorphic to $HN/N$.
Best Answer
Note that $|S_4|=4!=24$. Thus, $|G/N|=24$. Then, by Lagrange's theorem, $|G/N|=[G:N]$ and so $|G|=|G/N|\cdot|N|=24\cdot5=120$.
Now, consider the subgroup $K=\{(),(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\}\leq S_4\cong G/N$. Then $K\triangleleft G/N$, $|K|=4$, and $[G/N:K]=6$ (again by Lagrange's theorem). Thus, by the correspondence theorem, $K$ corresponds to a normal subgroup $H\triangleleft G$ such that $[G:H]=6$, and hence (by Lagrange's theorem) $|H|=120/6=20$.
Finally, since $G/N\cong S_4$ and $S_4$ has four subgroups of order $3$ ($\langle(1,2,3)\rangle,\langle(1,2,4)\rangle,\langle(2,3,4\rangle,\langle(1,3,4)\rangle$), none of which are normal in $S_4$, the desired result follows by application of the correspondence theorem.