I am having some trouble understanding the proof of Scott's core theorem as presented by Rubinstein and Swarup (link, requires academic access).
Scott's core theorem states the following: if $M$ is a topological $3$-manifold with finitely generated fundamental group, then any compact, incompressible, $3$-submanifold $K$ is contained in a core $K'$ of $M$. By incompressible we mean that the frontier $\overline{K} \cap \overline{M - K}$ contains no nontrivial loops that contract trivially in $M$. By core we mean that the inclusion of $K'$ into $M$ induces an isomorphism of their fundamental groups.
In Rubinstein/Swarup, they first prove the theorem in the special case that the frontier of $K$ has finitely many components, and that the fundamental group of each component of $K$ is indecomposable and not infinite cyclic. They then use this special case to prove the relative core theorem, which they then use to prove the general case. The proof of the relative core theorem is where I am getting confused.
THEOREM (relative core theorem): If $M$ is a $3$-manifold with finitely generated fundamental group, then any compact submanifold of the boundary $C \subset \partial M$ is contained in a core of $M$ whose intersection with the boundary is $C$.
For the moment, let's assume that $C$ is connected and incompressible. If I understand this case, it will be easy to generalize. To prove the relative core theorem, they consider small annular collars $A_i$ in $C$ around each component of $\partial C$. They then let $T$ be a topological torus with a hole cut out, and glue $T \times [0, 1]$ to each annulus $A_i$ along the hole. The resulting space $C'$ looks like $C$ with a thickened torus sprouting off of each boundary component $\partial C$. The claim is that by the special case of Scott's core theorem, this space $C'$ is contained in a core of $M'$, which is the manifold $M$ together with the glued tori.
I am having trouble understanding why the special case of the core theorem applies. $C'$ is compact, its frontier has finitely many components (in fact a single component, which is homeomorphic to $C$), but I do not see why its fundamental group is indecomposable and not infinite cyclic. If $C$ is simply connected, for example, then I would think that $C'$ should be homeomorphic to a thickened torus, which has decomposable fundamental group $\mathbb{Z} * \mathbb{Z}.$ In fact, I can't see why we should glue tori at all instead of gluing spherical caps; if I glued hemispheres to each annulus, I would think $C'$ should always be homeomorphic to a thickened $S^2$, which has trivial fundamental group and thus satisfies the special case of the core theorem.
These are my two chief confusions: (i) Why should the surface obtained by gluing tori to $C$ have an indecomposable, not-infinite-cyclic fundamental group? (ii) What is the point of gluing tori? Why not just glue hemispheres?
Any help?
EDIT: I have attached the following image of definitions from the relevant paper in response to some comments.
