Consider a group epimorphism $\varphi_1:G\to\mathbb{Z}$. Is there any way to understand the kernel of the map $\varphi_2:G\times G\to\mathbb{Z}$ given by $\varphi_2(g,h)=\varphi_1(g)-\varphi_1(h)=\varphi(gh^{-1})$ in terms of $\ker(\varphi_1)$?
Obviously $\ker(\varphi_1)\times \ker(\varphi_1)\subset\ker(\varphi_2)$ and $\Delta=\lbrace (g,g)\mid g\in G\rbrace\subset\ker(\varphi_2)$ but there could be more elements since $\varphi_1$ need not be injective.
Is there any way to express $\ker(\varphi_2)$ in a closed form? (Maybe a semidirect product or something like that)
Note that $\ker(\varphi_2)$ is a subdirect product of $G\times G$, since the projections onto each component are surjective. The subdirect products of $A\times B$ are classified by Goursat's Lemma: they are precisely the graphs of isomorphisms $A/N\cong B/M$ for normal subgroups $N$ of $A$ and $M$ of $B$.
Here, you get the graph of the identity map $\frac{G}{\ker(\varphi_1)}\cong\frac{G}{\ker(\varphi_1)}$. Thus, $$\ker(\varphi_2) = \bigcup_{g\in G} gN\times gN,$$ where $N=\ker(\varphi_1)$.
In particular:
Theorem. The elements of $\ker(\varphi_2)$ are the elements of the form $(g,gn)$ with $n\in N=\ker(\varphi_1)$.
Proof. All elements of the form $(g,gn)$ lie in $\ker(\varphi_2)$, since $\varphi_1(g)-\varphi_1(gn) = \varphi_1(g)-\varphi_1(g) = 0$.
Conversely, if $(x,y)\in\ker(\varphi_2)$, then $\varphi_1(x)=\varphi_1(y)$, so $xN=yN$. Thus, $y=xn$ for some $n\in N$, and $(x,y)=(x,xn)$. $\Box$
Note that this holds for any morphism into an abelian group $f\colon G\to A$, whether it is surjective or not.