The (finite) "girthwidth" is always 1 so this is not going to be that useful (as defined here anyway). First some notation: Let a $g$-decomposition of a graph $G$ consist of a graph $H$ with girth at least $g$ and an assignment to each node of $H$ of a set of vertices as above, the width of such a decomposition be one less than the maximum number of vertices assigned to any node of $H$, and $w_g(G)$ be the minimum width of any $g$-decomposition of $G$. Then $w_3(G)$ is the minimum width where $H$ is completely unrestricted. So $w_3(G)=1$ as we can take $H$ to be a complete graph and put the endpoints of each edge in their own node. Since the girth of a tree is infinite a tree decomposition is an $\infty$-decomposition and the tree width is $w_{\infty}$. Actually these are the only widths that matter since $w_3=w_b \le w_{\infty}$ for any finite $3 \le b$: If $a \lt b \le \infty$ then $w_a \le w_b$ simply by the definitions. However if $a \lt b \lt \infty$ then also $w_b \le w_a$ since we can turn an $a$-decomposition into a $2a$-decomposition by putting a new node in the middle of each edge and assigning it the same vertex set as one of its neighbors.

Maybe one could restrict to $H$ being a planar graph although $w_3(G)$ would still be 1 for planar $G$. I wonder when the "cycle-width" is less than the path-width. If one demands that no two nodes get the same vertex set (which is no restriction for tree width) it would get harder (and sometimes impossible if we forbid empty nodes)

A use of tree-width is that many hard (NP complete, say) problems become easy (say polynomial time) when restricted to bounded tree-width. A variant width could be justified merely by its own intrinsic appeal, but would be more motivated if it had similar applications.