Does replacing each vertex of $G$ by $H$ increase treewidth of $G$ by at most $\Delta(G)$? I am asking this question from the context of parameter preserving reductions which has implications for kernelization (See Theorem 18 of [1] for an example). For simplicity, here I am assuming that the host graph $G$ is regular. It seems to me that the game theoretic definition of treewidth is more useful here (See Section 2 of [2]).
Let $G$ be a regular graph with maximum degree $\Delta$. Let $H$ be a graph with $\Delta$ marked vertices. Let $G^*$ be the graph obtained from $G$ by replacing each vertex $v$ of $G$ by a copy of $H$ (say, each marked vertex of $H$ takes one neighbor of $v$ each).
Can we say that treewidth of $G^*$ is at most $\max\{tw(G)+\Delta,tw(H)+\Delta\}?$
(Let's use $tw(J)$ to denote the treewidth of a graph $J$.
For the special case $tw(H)\leq tw(G)$, we get the question in the title).
In [1], they consider the case when $H$ is sunlet$_6$ (i.e., $C_6$ plus a leaf attached to each vertex). In that paper, it is claimed that $tw(G^*)\leq tw(G)+2$. Is this claim correct?
(Disclaimer: it is possible that their claim is correct, but my idea of the generalization given as the first question is wrong).
The presentation in [1] gave me the feeling that the answer to the first question is yes. But, I couldn't find a reasonable explanation. I would like to know whether I am overlooking something well-known here.
Similar Situations
We can say the following.

If one vertex $v$ of $G$ alone is replaced by a copy of $H$, then the treewidth of the resultant graph is at most $\max\{tw(G)+\Delta,tw(H)+\Delta\}$.

Idea (in terms of cops and robber game): In an optimal cop strategy of $G$, whenever there is a cop at $v$, place $\Delta$ cops at marked vertices of $H$. Also, choose one of those points, make a new branch there which raids $H$. This forms a cops strategy for the resultant graph.
(Idea in terms of tree-decomposition: In the tree-decomposition of $G$, in each bag containing $v$, replace $v$ by marked vertices of $H$. Also, choose one such bag $B$ and append a new branch there that 'raids' $H$; in other words, take the tree-decomposition of $H$, add marked vertices of $H$ to each bag, and append this as a branch to the vertex associated with the bag $B$.)

Let $K$ be a graph with two marked vertices. Let $G'$ be the graph obtained from $G$ by replacing each edge $e$ of $G$ by a copy of $K$ (say, each marked vertex of $K$ takes one incident vertex of $e$ each). Then, the treewidth of $G'$ is at most $\max\{tw(G)+\Delta,tw(K)+\Delta\}$.

Idea (in terms of cops and robber game): Consider an optimal cops search strategy. For each edge $e=uv$ of $G$, choose a point in the strategy when edge $e$ is searched (i.e., there is a cop at $u$ and a cop at $v$) and raid $H$ at that point. This forms a cops strategy for $G'$.
Thank you.
References
[1] Lauri, Juho; Mitillos, Christodoulos, Complexity of fall coloring for restricted graph classes,  ZBL07173544.
[2] Langer, Alexander; Reidl, Felix; Rossmanith, Peter; Sikdar, Somnath, Practical algorithms for MSO model-checking on tree-decomposable graphs, Comput. Sci. Rev. 13-14, 39-74 (2014). ZBL1302.68184.
Let $G$ be a complete graph with $n$ vertices and $H$ be a complete graph with $n-1$ vertices, all marked. We replace each vertex of $G$ with $H$ to produce a graph $G^*$ with $n(n-1)$ vertices that is $n$-regular. We claim that the treewidth of $G^*$ is at least $n (n-1)/12$, which falsifies the conjecture. Proof:
Suppose that the treewidth of $G^*$ is less than $n(n-1)/12$. Therefore by standard results [1] its vertices can be partitioned to sets $A$, $S$, $B$ such that there are no edges between $A$ and $B$, and $|S| = n(n-1)/12$, $|A| \ge n(n-1)/4$, $|B| \ge n(n-1)/4$, and $|A| + |B| = n(n-1)11/12$.
We call the cliques $H$ in $G^*$ corresponding to vertices of $G$ clusters. The graph $G^*$ has $n$ clusters. Note that each cluster has vertices from at most one of $A$ or $B$. Let $A_f$ be the fraction such that $n A_f$ clusters contain vertices from $A$ and $B_f$ the fraction such that $n B_f$ clusters contain vertices from $B$.
Now, $A_f \ge 1/4$, $B_f \ge 1/4$, and $A_f + B_f \ge 11/12$.
Notice that if there is a cluster $v$ that contains vertices from $A$ and a cluster $u$ that contains vertices from $B$ then these clusters are connected by an edge $\{v_i, u_j\}$, both vertices in it unique to that pair of clusters, and at least one of $v_i$ and $u_j$ should be in $S$.
Therefore we get that $|S| \ge n^2 A_f B_f$, which we can solve with our constraints to a minimum value $|S| \ge n^2 A_f B_f \ge n^2 8/12 \cdot 1/4 = n^2 1/6$, which is a contradiction to the assumed size of $S$.
[1] http://parameterized-algorithms.mimuw.edu.pl/parameterized-algorithms.pdf