The context for this question comes from this arxiv preprint. Specifically, a remark in the final proof of the paper. To make the question more self-contained, I'll phrase this question in a slightly more general setting.
Fix a category $A$ and a combinatorial model category $\mathcal{M}$ where all objects are cofibrant [1]. Given a pair objects $X : \mathcal{M}$ and $a : A$, we define $a \otimes X$ to be the left Kan extension of $X$ along the inclusion $\{a\} \to A$. Explicitly then,
$$ (a \otimes X)(a') = \hom(a,a') \cdot X $$
The question: Endowing $\mathbf{Fun}(A,\mathcal{M})$ with the projective model structure, can an arbitrary functor $F : A \to \mathcal{M}$ be built out of an (iterated) homotopy colimit of $a \otimes X$s?
As a first step, we can use the classical Yoneda formula to realize $F$ as $\int^{a : A} a \otimes F(a)$, so $F$ can certainly built as the coequalizer of coproducts of presheaves of the form $a \otimes X$. I'm unsure how to argue that this is the homotopy colimit of some functor. It seems natural enough to extend this coequalizer formula to a simplicial object:
$$ G(n) = \coprod_{a_0 \to \dots \to a_n} a_0 \otimes F(a_n) $$
This then appears to be a cofibrant diagram with respect to the Reedy model structure on simplicial objects of $\mathbf{Fun}(A, \mathcal{M})$ whose colimit is $F$. However, my understanding is that this alone does not imply that $\mathsf{hocolim}\,G = F$ since $\Delta : \mathbf{Fun}(A, \mathcal{M}) \to \mathbf{Fun}(\Delta^\mathsf{op}, \mathbf{Fun}(A, \mathcal{M}))$ is not necessarily a right Quillen functor.
Is it still the case that $\mathsf{hocolim}\,G = F$? Or do I need to search for a different $G$ entirely? I asked a similar question on a different forum and was adviced to use the Bousfield-Kan formula. This certainly appears to be quite related but I'm not quite sure how to join the pieces together.
[1]: For the actual example, $\mathcal{M}$ is $\mathbf{sSet}^\sharp_{/B}$ with the Cartesian model structure.
