Apologies in advance for the vague question.
Suppose I have a homomorphism $\phi$ from the fundamental group of a surface of genus g into a free product of groups, $A\ast B$. I want to represent the free product as a space, for example $K(A,1)\vee K(B,1)$, and $\phi$ as a continuous map $f: \Sigma_g\rightarrow K(A,1)\vee K(B,1)$. Can I say that the pre-image of the wedge point, $f^{-1}(x)$, is just a collection of disjoint embedded loops in $\Sigma_g$ (or at least that this is true for some map in the homotopy class of $f$)? Can I also somehow say that I can split the surface into two submanifolds, $T_A$ and $T_B$, such that $f(T_A)\subset K(A,1)$ and $f(T_B)\subset K(B,1)$?
For context I'm trying to understand the second paragraph in Stallings' proof of Theorem 2 in his paper "How not to prove the Poincare Conjecture".
I don't know whether, in what you are reading, it matters what choice one makes of $K(A,1)$ and $K(B,1)$ and their base points. I sort of doubt that it matters, and with one particular choice there's a pretty straightforward answer.
Let's start with any choices of $K_A$ for $K(A,1)$ and $K_B$ for $K(B,2)$. Let $X$ be obtained from the disjoint union of $K_A$ and $K_B$ by attaching an edge $E$, identifying one endpoint of $E$ with the base point of $K_A$, and the opposite end point with the base point of $K_B$. Notice: $X = K'_A \vee K'_B$ where $K'_A$ is obtained by attaching half of $E$ to $K_A$, and $K'_B$ is obtained by attaching the other half of $E$ to $K_B$, so $K'_A,K'_B$ deformation retract to $K_A,K_B$, hence $K'_A,K'_B$ are still Eilenberg-Maclane spaces.
Since the midpoint $x \in E$ is a 1-manifold point, any continuous map $\Sigma_g \to X$ may be perturbed to obtain a map $f : \Sigma_g \to X$ which is transverse to $x$. And now all the conclusions that you desire in your post do actually hold: $f^{-1}(x)$ is a collection of disjoint embedded loops, subdividing $\Sigma_g$ into two subsurfaces, one mapping to $K'_A$ and the other to $K'_B$.