Let $\phi:G\to G' , H\leq G$.
Prove $\phi^{-1}(\phi(H))=\langle H,\ker\phi \rangle$
Proof:
First, $\phi^{-1}(\phi(H)) \subseteq \langle H,\ker\phi \rangle$
Case 1: $H\subseteq \ker\phi\implies \phi^{-1}(\phi(H))=\ker\phi \in \langle \ker\phi \rangle\subseteq \langle H,\ker\phi \rangle$
Case 2: $H\not\subseteq \ker\phi \implies \forall h\in H\setminus \ker\phi $ exists $\phi^{-1}\phi(h)=h\in\langle H \rangle \subseteq \langle H,\ker\phi \rangle.$
Otherwise, $h\in \ker\phi,$ hence $\phi^{-1}(\phi(h))=\phi^{-1}(e_{G'})\in \ker \phi \subseteq\langle H,\ker\phi \rangle.$
Then, $\phi^{-1}(\phi(H)) \subseteq \langle H,\ker\phi \rangle$.
Second, $ \langle H,\ker\phi \rangle \subseteq \phi^{-1}(\phi(H))$
I have no idea how to approach the second direction, any help is welcome thanks!
Let $h\in H$. Then $\phi(h)\in \phi(H)$ and therefore $h\in \phi^{-1}(\phi(H))$. We have $H\subseteq \phi^{-1}(\phi(H))$.
Next, let $x\in \ker \phi$. Then $\phi(x)=e_{G'}\in \phi(H)$ and therefore $x\in \phi^{-1}(\phi(H))$. We have $\ker \phi\subseteq \phi^{-1}(\phi(H))$ and we can conclude that $\langle H, \ker \phi\rangle\subseteq \phi^{-1}(\phi(H))$.
Anyway, I think your Case 2 is not correct because $\phi^{-1}(\phi(h))$ is not necessarily equal to $h$ for $h\in H\setminus \ker\phi$.
To show that $\phi^{-1}(\phi(H)) \subseteq \langle H, \ker \phi\rangle$, let $x\in \phi^{-1}(\phi(H))$. Then $\phi(x)=\phi(h)$ for some $h\in H$. This implies that $\phi(xh^{-1})=\phi(x)\phi(h)^{-1}=e_{G'}$ and hence $xh^{-1}\in \ker\phi$. Therefore $xh^{-1}=y$ for some $y\in \ker\phi$ and we have $x=yh\in \langle H, \ker \phi\rangle$.