# Show that the lie brackets $ad(X_e)(Y_e)=[X,Y]_e$ coincide

I have two descriptions of the lie algebra of a lie group. Both come with their version of the lie bracket

1. $$\mathfrak{g}=T_eG$$ as the tangent space at the identity comes with the lie bracket $$ad(X_e)(Y_e)$$.
2. $$\mathfrak{g}=\mathfrak{X}^{inv}(G)$$ as the left invariant vector fields comes with the lie bracket $$[X,Y]$$.

After identifying both descriptions of the lie algebras, I wish to show their brackets coincide: $$ad(X_e)(Y_e)=[X,Y]_e$$. These notes explain this but I do not follow his second to last step: How do I arrive at the sum

$$\frac{\partial^2}{\partial s\partial t}|_{0}f(expt(tX)exp(sY))+\frac{\partial^2}{\partial s\partial t}|_{0}f(expt(sY)exp(-tX))$$?

I tried to see if it follows from some Leibniz rule or multivariable calculus, but I fail to reproduce his result. I have seen a different proof of this result in Lee where it uses the pullback of the flow, but I fail to see how this applies here.

Morally, it is just an instance of the Leibniz rule where, computing $$\frac{\partial}{\partial t}|_{t=0}(e^{tX}e^{sY}e^{-tX})$$, you apply $$\frac{\partial}{\partial t}|_{t=0}$$ first to the factor $$\exp(tX)$$ and then to the factor $$\exp(-tX)$$.
Adding more detail, let us recall the canonical isomorphism of vector bundles, covering the identity map $$\operatorname{id}_{G\times G}$$, $$T(G\times G)\overset{\sim}{\longrightarrow}TG\times TG$$ which maps $$\frac{\partial}{\partial t}|_{t=0}(\gamma_1(t),\gamma_2(t))$$ to $$(\frac{\partial}{\partial t}|_{t=0}\gamma_1(t),\frac{\partial}{\partial t}|_{t=0}\gamma_2(t))$$, for any pair of smooth curves $$\gamma_1,\gamma_2\colon I\to G$$.
Understanding the previous isomorphism,one gets, in particular $$\tfrac{\partial}{\partial t}|_{t=0}(e^{tX}e^{sY},e^{-tX})=(\tfrac{\partial}{\partial t}|_{t=0}e^{tX}e^{sY},\tfrac{\partial}{\partial t}|_{t=0}e^{-tX})=(\tfrac{\partial}{\partial t}|_{t=0}e^{tX}e^{sY},0_e)+(0_{e^{sY}},\tfrac{\partial}{\partial t}|_{t=0}e^{-tX})=\tfrac{\partial}{\partial t}|_{t=0}(e^{tX}e^{sY},e)+\tfrac{\partial}{\partial t}|_{t=0}(e^{sY},e^{-tX}).$$
Finally, the latter allows to compute $$\tfrac{\partial}{\partial t}|_{t=0}f(e^{tX}e^{sY}e^{-tX})=\tfrac{\partial}{\partial t}|_{t=0}f(m(e^{tX}e^{sY},e^{-tX}))=(f\circ m)_*\tfrac{\partial}{\partial t}|_{t=0}(e^{tX}e^{sY},e^{-tX})=(f\circ m)_*\tfrac{\partial}{\partial t}|_{t=0}(e^{tX}e^{sY},e)+(f\circ m)_*\tfrac{\partial}{\partial t}|_{t=0}(e^{sY},e^{-tX})=\tfrac{\partial}{\partial t}|_{t=0}f(m(e^{tX}e^{sY},e))+\tfrac{\partial}{\partial t}|_{t=0}f(m(e^{sY},e^{-tX}))=\tfrac{\partial}{\partial t}|_{t=0}f(e^{tX}e^{sY})+\tfrac{\partial}{\partial t}|_{t=0}f(e^{sY}e^{-tX}),$$ where $$m\colon G\times G\to G$$ denotes the group multiplication.