I don't understand a nuance in Theorem 2.4.4.10. of Kerodon. To not burden the question with exposition, I will only provide information for the relevant part.
Fix a natural number $m$. For a simplicial set $S$, let $E(S,m)$ denote the set of all pairs $(\sigma, I)$ where $\sigma\colon \Delta^n\to S$ is a non-degenerate $n$-simplex of $S$ for $n > 0$ and $I$ is a chain $I_m \subseteq I_{m-1} \subseteq \dots \subseteq I_1 \subseteq I_0$ of subsets of $[n]$ such that $I_0 = [n]$ and $I_m = \{0,n\}$. We then define a directed graph (or a quiver) $G(S)$ as follows: vertices of $G(S)$ are precisely the vertices of $S$, that is, its $0$-simplices, and edges are the elements of the set $E(S,m)$ where an edge $(\sigma, I)$ has a source $\sigma_0(\langle 0 \rangle)$ and a target $\sigma_0(\langle n \rangle)$ (here by $\langle i \rangle$ I mean the unique map $[0]\to [n]$ which takes the value $i$).
Let
$\require{AMScd}$ \begin{CD} S @>>> T\\ @VVV @VVV\\ S' @>>> T' \end{CD}
be a pushout of simplicial sets with horizontal maps being monomorphisms. Jacob Lurie, the author of the book, claims that there there is pushout
$\require{AMScd}$ \begin{CD} G_0(S) @>>> G_0(T)\\ @VVV @VVV\\ G(S') @>>> G(T') \end{CD}
where the vertices of $G_0(S)$ coincide with that of $G(S)$, while it has no edges, and the vertices of $G_0(T)$ also coincide with the vertices of $G(T)$, while the set of its edges is $\mathrm{Edge}(G(T))\setminus\mathrm{Edge}(G(S))$.
The obvious guess is that it is a composition of pushouts
$\require{AMScd}$ \begin{CD} G_0(S) @>>> G_0(T)\\ @VVV @VVV\\ G(S) @>>> G(T) \\ @VVV @VVV\\ G(S') @>>> G(T') \end{CD}
However, the problem is that $G$ is not a functor (at least not a functor from the full category of simplicial sets). Indeed, G can only act in the obvious way on morphisms which are monomorphisms, so that they carry non-degenerate simplicies to non-degenerate simplices. In particular, I don't see a natural way to define a function $E(T,m)\setminus E(S,m)\to E(T',m)$ which is induced by $T\to T'$.
Come to think of it, this follows from the concrete description of pushouts in $\mathsf{Set}_{\Delta}$ (or, more generally, in presheaf categories).
Let
$\require{AMScd}$ \begin{CD} S @>>> T\\ @VVV @VVV\\ S' @>>> T' \end{CD}
is a pushout in $\mathsf{Set_{\Delta}}$. Denote $T\to T'$ by $q$ and $S\to S'$ by $f$ (for simplicity, assume the the horizontal maps are literally inclusions). Then $T'_n$ may be realized as the quotient $(S'_n\sqcup T_n)/{\sim}$ where $\sim$ is the equivalence relation generated by pairs $((0,s'),(1,t))$ such that $t \in S_n$ and $s' = f_n(t)$. Thus, if $t$ doesn't lie in $S_n$, $(1,t)$ is not equivalent to anything other than itself.
On the other hand, $q$ maps $t$ to the equivalence class $[(1,t)]$. Assume that $t \in T_n\setminus S_n$ is non-degenerate in $T$. Since $t$ is non-degenerate, for $[(1,t)]$ to be degenerate in $T'$ it must be equivalent to either $(0,s')$ where $s'$ is degenerate in $S'_n$ or to $(1,t')$ where $t'$ is a degenerate element of $T$. But this is impossible by the above.