Let $G = H_1 \ast_K H_2$ be an amalgamated free product of two groups such that $K$ has finite index in both $H_1$ and $H_2$. Let $G'$ be a finite index subgroup of $G$. Does it follow that $G'$ splits as a non-trivial amalgamated free product or as an HNN-extension? Does it follow in the case we assume $G'$ is normal in $G$?
I certainly think the answer to all of this is "no", but I do not see a nice counterexample (even for a subgroup of an amalgam of the form $\mathbf{Z} \ast_{\mathbf{Z}} \mathbf{Z}$). When the amalgam is of finite groups, then Stallings' theorem tells us that the answer is yes, however.
I think the only real way to approach this problem is using the theory of group actions on trees, otherwise known as Bass-Serre Theory, and so I'll write my answer with Bass-Serre theory as a prerequisite, although my list of bullet points kinda/sorta says everything that needs to be said to solve this problem.
Let me state the unstated nontriviality property in the setting of your post, namely that the given injections from $K$ into both $G$ and $H$ have images of index $\ge 2$ (without that property the theory becomes trivial: every group splits over every subgroup).
The answer to your main question is "yes": denoting your group as $\Gamma = G *_K H$, every finite index subgroup $\Gamma' < \Gamma$ splits nontrivially as an amalgamated free product or HNN extension.
For the proof, let $T$ be the Bass-Serre tree of the splitting:
By restriction, the subgroup $\Gamma'$ also acts on $T$ by simplicial isomorphisms. And this action is also minimal: if it were not, then the action of $\Gamma'$ on $T$ would have a proper minimal subtree $T' \subset T$, and each component of $T \setminus T'$ would be an infinite subtree of $T$, from which it follows that $\Gamma'$ would have infinite index in $\Gamma$.
Another property of the action of $\Gamma'$ on $T$ is that it has a fundamental domain which is a finite subtree $\tau \subset T$; one can show in general that there is some subtree $\tau$ which is a fundamental domain, but if that subtree were infinite then, again, $\Gamma'$ would be of infinite index.
One can now derive a "finite graph of groups" presentation of $\Gamma'$, by taking the quotient graph $T / \Gamma'$; equivalently, it is a finite graph obtained from the chosen fundamental domain $\tau$ by identifying two vertices of $\tau$ when they are in the same $\Gamma'$ orbit. One uses vertex and edge stabilizers, and elements of $\Gamma'$ taking one vertex of $\tau$ to another vertex in the same orbit, for the purpose of carefully labelling the edges and vertices of the quotient graph, and the edge-group-to-vertex-groups injections, in order to get a graph of groups.
And, finally, your desired splitting of $\Gamma'$ comes from picking one edge of the quotient graph $T/\Gamma'$ and collapsing all the rest: if that edge is separating then one gets an amalgamated free product presentation; whereas if that edge is nonseparating then one gets an HNN amalgamation.
But the answer to the subquestion in your comment is "no", the splitting of $\Gamma'$ need not have the form that you asked about. For instance the rank $2$ free group $\mathbb Z * \mathbb Z$ contains a rank $3$ free group $\mathbb Z * \mathbb Z * \mathbb Z$ with index $2$, and there's just no way to write that as the free product of two finite index subgroups of $\mathbb Z$. What one can do is to use the Bass-Serre theory to list all of the finite index subgroups of an amalagamated free product, expressed as finite graphs of groups using finite index subgroups of the given vertex groups and edge groups, in a similar fashion to how one uses covering graphs to list all of the finite index subgroups of a free group.