Let $G = ( \Bbb Z \times \Bbb Z , +)$ and $H = ( \Bbb Z, +).$ Also define $A = \{(x,x) \mid x \in \Bbb Z \}$. Let $f\colon G \to H,\, f(a,b) = a-b$. Show that $G / A \cong H$.
Clearly $f$ is a homomorphism and thus by the homomorphism theorem $G / \ker(f) \cong \operatorname{im}(f)$, with the map $F: G / \ker(f) \to \operatorname{im}(f), F(gA)=f(g)$.
So isn’t the isomorphism $G / A \cong H $ immediate when I notice that $A$ is actually $\ker(f)$? Since for any $a = (x,x) \in A$ I have that $f(a) = 0 = e_H$?
I would be a bit careful about the statement "cleary $f$ is a homomorphism"; this heavily depends on the codomain $H$ being abelian.
Putting this part aside you are right. The isomorphism theorem tells you that $G/\ker f\cong\operatorname{im}f$ so it only remains to determine $\ker f$ and $\operatorname{im}f$ as $A$ and $H$, respectively.
The kernel part is immediate since $a-b=0$ iff $a=b$. What you noted, however, is only one inclusion, namely $A\subseteq\ker f$, while you need the reverse inclusion too. The image part is likewise straightforward since $G=H^2$.
To summarize: your overall idea is correct but some details could be a bit more precise.