Why is this map from Leray spectral sequence zero?

I believe Tag 0BUT implies something stronger, namely that $R^{1}f_{\ast}\mathcal{O}_{X}^{\ast} = 0$. One can see this as follows: Let $\mathcal{F}$ be the presheaf on $Y$ given by $V \mapsto H^{1}(f^{-1}(V),\mathcal{O}_{X}^{\ast})$. Then $R^{1}f_{\ast}\mathcal{O}_{X}^{\ast}$ is the sheafification of $\mathcal{F}$. By Tag 0BUT, for any open subset $V \subseteq Y$ and $s \in \Gamma(V,\mathcal{F})$, there is an open cover $V = \bigcup_{j \in J} V_{j}$ such that $s|_{V_{j}} = 0$ for each $j \in J$. This means that the universal separated presheaf associated to $\mathcal{F}$ is $0$, so the sheafification is $0$ as well.

Edit : I assumed that $\mathcal O^*_X$ is quasi-coherent as $\mathcal O_X$-module which is wrong. However I think the argument should still holds since proper base change and Grothendieck theorem are valid in greater generality (i.e for sheaves of abelian groups).

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If you can use proper base change, then for any sheaf $F \in \mathcal O_X \text{-mod}$ you obtain $(R^1f_*F)_y \cong H^1(f^{-1}(y), F_{|f^{-1}(y)})$ where if $i : Z \to X$ is the inclusion of a closed subset then $F_{|Z}$ means $i^* F$. Since $f$ is finite $\dim f^{-1}(y) = 0$ and so by Grothendieck's vanishing (see here) we obtain $H^1(f^{-1}(y), F_{|f^{-1}(y)}) = 0$.