Let $G$ be a finite group and $\phi : G \to{\rm Aut}(F_2)$ be a homomorphism. For example, we have $\phi : \mathbb{Z}/2 \to{\rm Aut}(F_2)$ that switches the generators. What can I do to see if the semidirect product $F_2 \rtimes_\phi G$ is isomorphic to $F_2 \times G$ or prove that it is not? More generally, how do I see if for two different $\phi_1$ and $\phi_2$, the results are isomorphic groups?
Furthermore, I would love to know how to give an example of an extension of $F_2$ with finite quotient $G$ (ideally cyclic) that does not arise form this semidirect product construction.
If $\Gamma$ is your example $\mathbb Z_2\rtimes_{\varphi(\bar 1)} F_2$ you are trying to extend $F_2$ by $\mathbb Z_2$ which means that you are trying to complete a short exact sequence $$0\to F_2\to\Gamma\to \mathbb Z_2\to0.$$ It is known that the extensions $\Gamma$ are controlled by the morphisms $$\mathbb Z_2\to{\rm Out}(F_2)$$ where ${\rm Out}(F_2)={\rm Aut}(F_2)/{\rm Inn}(F_2)$. Happily also ${\rm Out}(F_2)=GL_2(\mathbb Z)$.
So, it is enough to set $\varphi:\mathbb Z_2\to{\rm Out}(F_2)$ by telling where the generator $\bar 1$ of $\mathbb Z_2$ goes to tell extensions in this settings.
If $\varphi(\bar 1)=1\!\!1$ is the $GL_2(\mathbb Z)$'s identity then $\Gamma=\mathbb Z_2\rtimes_{\varphi(\bar 1)} F_2=\mathbb Z_2\times F_2$. This has only one element, $(\bar 1,1)$, of order two.
If $\varphi(\bar 1)=\gamma$ is an order two element in $GL_2(\mathbb Z)$ then $\Gamma=\mathbb Z_2\rtimes_{\gamma} F_2$ has another element of order two which is $(\bar 0, 1)$, apart from $(\bar 1,1)$.