Let $X$, $Y$, $V$ be matrices.
$$e^{\tilde{V}}=e^{X}e^{V}e^{Y}$$
How to use the Baker–Campbell–Hausdorff Formula to prove the following identity?
$$\tilde{V}=V+\frac{1}{2}[V,Y-X]+\mathrm{ad}_{V/2}\coth(\mathrm{ad}_{V/2})(X+Y)+ ...,$$ where $\mathrm{ad}_{V}\cdot X=[V,X]$ is the Lie derivative.
As indicated in the comment, you really wish to show $$e^{\tilde{V}}=e^{X}e^{V}e^{Y}\equiv e^W e^Y\\ \tilde{V}=V+\frac{1}{2}[V,Y-X]+ \mathrm{ad}_{V/2}\coth(\mathrm{ad}_{V/2})(X+Y)+ O(X^2,Y^2,XY),$$ that is, all orders in V but only linear in X,Y. (You can tell this must be so from special cases X=0 or Y=-X.)
We'll use the standard partial CBH formulas from WP, $$ Z= \log (e^X e^Y) = X+\frac{ \mathrm{ad}_{X}}{1-e^{-\mathrm{ad}_{X}}}~Y +O(Y^2)= X+ \mathrm{ad}_{X/2}(1+\coth\mathrm{ad}_{X/2} )~Y +O(Y^2)= \\ = Y- \frac{ \mathrm{ad}_{Y}}{1-e^{\mathrm{ad}_{Y}}}~X +O(X^2) , $$ the second one following from the cornerstone CBH symmetry $Z(Y,X)=-Z(-X,-Y) $. We'll use this second one first, and then the first one.
We thus seek $$ \tilde V= \log ( e^{V-\frac{ \mathrm{ad}_{V}}{1-e^{\mathrm{ad}_{V}}}~X +O(X^2) } ~~e^Y ) = W + \frac{ \mathrm{ad}_{W}}{1-e^{-\mathrm{ad}_{W}}}~Y +O(Y^2). $$
Now, crucially, $ \mathrm{ad}_{W} \to \mathrm{ad}_{V}$ to linear order in X,Y, so that $$ \tilde V = V-\frac{ \mathrm{ad}_{V}}{1-e^{\mathrm{ad}_{V}}}~X +O(X^2)+ \frac{ \mathrm{ad}_{V}}{1-e^{-\mathrm{ad}_{V}}}~Y +O(Y^2) +O(XY) \\ = V + \frac{1}{2} \mathrm{ad}_{V} (Y-X) + \left ( \frac{ \mathrm{ad}_{V}}{1-e^{-\mathrm{ad}_{V}}} -\mathrm{ad}_{V/2} \right )~Y +\left (-\frac{ \mathrm{ad}_{V}}{1-e^{\mathrm{ad}_{V}}}+\mathrm{ad}_{V/2}\right )X +O(...) $$ $$ = V + \frac{1}{2} [V,Y-X] + \bbox[yellow]{\mathrm{ad}_{V/2} \coth ( \mathrm{ad}_{V/2}) ~( Y+X) +O(...)}. $$