The Verlinde formula gives an explicit formula for any finite dimensional simple Lie algebra $\mathfrak{g}$ an explicit formula for the dimension of conformal blocks for the associated WZW conformal field theory. Explicitly, we pick a smooth curve $X$ and weights $\lambda_i$ (or equivalently their Verma modules $M_{\lambda_i}$) attached to finitely many points $x_i\in X$. The space $V_X(\lambda,x)$ of conformal blocks should then be thought of as "placing the $M_{\lambda_i}$ at $x_i$, and take some version of global sections".
See e.g. Beauville's Conformal blocks, fusion rules and the Verlinde formula, which does it for all $\mathfrak{g}$ except type $F$. Corollary $9.8$ there gives $$\text{dim}V_X(\lambda,x)\ =\ |T_\ell|^{g-1}\sum_{\mu\in P_\ell}\text{Tr}_{V_\lambda}(\exp 2\pi i \frac{\mu+\rho}{\ell+h^\vee})\prod_{\alpha>0}\left|2\sin \pi\frac{(\alpha|\mu+\rho)}{\ell+h^\vee}\right|^{2-2g}. $$
However, a more modern point of view on conformal field theories and conformal blocks is to work inside the category of $\mathcal{D}$ modules on the Ran space $\text{Ran}X$ parametrising finite subsets of $X$. We can then interpret the conformal field theory as an algebra $\mathcal{V}$ in $\mathcal{D}-\text{Mod}(\text{Ran}X)^{\otimes^{ch}}$, and (I think) "$M_{\lambda_i}$ at $x_i$" as modules $\mathcal{M}_i$ for this algebra, and (I think) conformal blocks are just the global sections (i.e. de Rham cohomology) of their chiral tensor product, or something like that.
My question is: is it possible to prove the Verlinde formula in a "factorisation algebraic" way, e.g. mostly working on the Ran space? Is there a "factorisation algebraic" interpretation of the terms in the right hand side of the Verlinde formula?
