Realising Seifert-van Kampen in 2-complexes Suppose you have a group $G$ given by a finite presentation $\langle X; \mathbf{r}\rangle$, and also suppose you know that this group decomposes as a free product with amalgamation $A\ast_CB$ (with $A$, $B$ and $C$ all infinite, and I am happy to assume $C\cong\mathbb{Z}$ for simplicity).

Is there any way we can link the presentation complex $\mathcal{C_P}$ of $G$ (single vertex, edges labelled from $X$, 2-cells attached using $\mathbf{r}$) with the Seifert-van Kampen-esque 2-complex $\mathcal{C}_{SvK}$ underlying $A\ast_CB$?

My initial thought was to look at the cover associated to the amalgamating subgroup $C$, and ask if I get a complex with a subcomplex which satisfies Seifert-van Kampen. However, this doesn't seem to work even if I take the cover of the presentation complex of $A\ast_CB$.
 A: Hatcher in Algebraic Topology discusses "graphs of groups" at the end of section 1.B.  One example is the realization of an Eilenberg-MacLane space $K(A*_CB,1)$ from a $K(A,1)$, a $K(B,1)$, and a $K(C,1)$.  The construction uses the fact that any homomorphism $\pi_1(X)\to A$ for a CW complex $X$ is induced by a map $X\to K(A,1)$, unique up to homotopy.  The two homomorphisms $C\to A$ and $C\to B$ correspond to a pair of maps $K(C,1)\to K(A,1)$ and $K(C,1)\to K(B,1)$ that can be used in a mapping-cylinder-like construction.  In particular, you take a mapping cylinder for each and then join them along the $K(C,1)$.
The two-skeleton of a $K(A*_CB,1)$ can be thought of as a presentation complex for this amalgamated product, what you were calling $\mathcal{C}_{SvK}$.  The cell structure of the Eilenberg-MacLane space can be arranged to have only a single $0$-cell, if you so choose.
An isomorphism $G\to A*_CB$ gives rise to a homotopy equivalence $K(G,1)\to K(A*_CB,1)$, which we can take to be cellular (the $n$-skeleton maps into the $n$-skeleton).  If you restrict to the two-skeleta, you get a map $f:\mathcal{C}_P\to\mathcal{C}_{SvK}$ (in your terminology) that induces an isomorphism on $\pi_1$.  It's worth mentioning that a $K(G,1)$ can be built by extending a presentation complex for $G$ using only $(n\geq 3)$-cells.
What follows is the beginnings of locating $C$ in $\mathcal{C}_P$.
Let's consider a complex for $\mathcal{C}_{SvK}$ from the mapping cylinder construction.  Call the subcomplexes from each group $\mathcal{C}_A$, $\mathcal{C}_B$, and $\mathcal{C}_C$, where there are some remaining $1$-cells and $2$-cells in the complement from attaching the $\mathcal{C}_C$ to either complex.
Now, let's suppose that each $2$-complex's $2$-cells' attaching maps are "taut," in that there is no backtracking or pausing and it is basically piecewise linear (the boundary of the $D^2$ is subdivided into intervals for how it is glued into the $1$-skeleton, just like the usual identification diagrams).  We can make the image of $f$ transverse to $\mathcal{C}_C$ in a sense.  By a homotopy of $f$, we can make the $0$-skeleton avoid $\mathcal{C}_C$, and instead land in the $0$-skeleta of $\mathcal{C}_A$ and $\mathcal{C}_B$.  Now consider the image of a $1$-cell of $\mathcal{C}_P$ through $f$, which we may assume is taut after a homotopy of $f$.  Since each $1$-cell has an incident $2$-cell from the mapping cylinder construction, we can homotope $f$ so that the $1$-cell crosses through $\mathcal{C}_C$ at a $0$-cell, without the path spending any time in $\mathcal{C}_C$.  Lastly, we can assume the images of $2$-cells are taut, and then  if a $2$-cell spends any time on a $2$-cell of $\mathcal{C}_C$, the boundary of that $2$-cell has an image on $\mathcal{C}_A$ and $\mathcal{C}_B$, so the map can be modified without changing the fact that $f_*$ is an isomorphism on $\pi_1$ (this amounts to homotoping the $\mathcal{C}_P\to K(A*_CB,1)$ map, which still had $3$-cells).  By another homotopy, we can make it so the $2$-cells contain entire $1$-cells of $\mathcal{C}_C$ if they intersect the interior of the $1$-cell at all.
Then, $f^{-1}(\mathcal{C}_C)$ consists of points of the $1$-skeleton (finitely many per $1$-cell) along with loops and arcs on the $2$-cells, with the arcs connecting these points.  There is a modification of $f$ that can remove these loops: since they bound disks, the image of the disk can be made to avoid the $\mathcal{C}_C$, essentially by a homotopy of the original $\mathcal{C}_P\to K(A*_CB,1)$.  If one could modify $f$ to make the preimage connected, then this would decompose $\mathcal{C}_C$ into two pieces, with the $\pi_1$-image of the common intersection lying in the image of $C$ in $A*_CB$.
Example. Let $\Sigma$ be a genus-$g$ closed surface, which is already a presentation complex of $\pi_1(\Sigma)$.  If $\pi_1(\Sigma)$ were represented as an amalgamated product, the preceding would give that there is a map $f:\Sigma\to \mathcal{C}_{SvK}$  with $f^{-1}(\mathcal{C}_C)$ being a collection of disjoint curves, none of which bound disks.  There are two kinds of curves: non-separating and separating, where the latter must be of the type that represent a connect-sum decomposition.  By some theory of non-separating curves on surfaces, since any closed loop must intersect transversely the separating curves an even number of times (since $\mathcal{C}_{SvK}-\mathcal{C}_C$ is two pieces), they must come in homotopic pairs, and so by a modification of $f$ we can eliminate the pairs one at a time while still giving an isomorphism on $\pi_1$. So, using the fact that there are at most $g-1$ pairwise non-homotopic separating curves that are not nullhomotopic: the $g=1$ case implies that $\mathbb{Z}^2$ is not a non-trivial amalgamated product, and the $g=2$ case implies every non-trivial amalgamated product comes from a connect sum of two tori (i.e., $\pi_1(\Sigma_2)=F_2*_{\mathbb{Z}}F_2$ with $1\mapsto aba^{-1}b^{-1}$ is the only nontrivial splitting, up to isomorphism).  I'm not sure what can be said about $g>2$, though! I couldn't figure out how to join separating curves together into one, if that is even possible.
