(this question uses physics conventions, which might differ by factors of $i$ from pure math conventions)
Suppose I have a Lie group $G$ with generators $T_i$. Their commutator is $$[T_i,T_j] = if_{ijk}T_k$$ with some structure constants $f_{ijk}$. Furthermore, the (left-) Lie derivative of any scalar function $g:G\to\mathbb{R}$ is defined as $$ \partial_ig(U) = \lim_{h\to0}\frac{1}{h}\left(g(e^{ihT_i}U)-g(U)\right) $$.
Question: what is the commutator $[\partial_i,\partial_j]=?$ It should intuitively be proportional to $f_{ijk}\partial_k$. But what is the correct sign? My sources (such as Eq. 5.3 in this paper) indicate it should be $-f_{ijk}\partial_k$. But my calculation comes out as $+f_{ijk}\partial_k$. Would appreciate if someone could show me my error.
My calculation: \begin{align} [\partial_i,\partial_j]g(U) &= \partial_i \lim_{h\to0}\frac{1}{h}\left(g(e^{ihT_j}U)-g(U)\right) - (i\leftrightarrow j) \\ &= \lim_{h\to0}\frac{1}{h^2}\left(g(e^{ihT_j}e^{ihT_i}U) - g(e^{ihT_j}U)-g(e^{ihT^i}U)+g(U)\right) - (i\leftrightarrow j) \\ &= \lim_{h\to0}\frac{1}{h^2}\Big(g(\underbrace{e^{ihT_j}e^{ihT_i}}_{=e^{ihT_i}e^{ihT_j}e^{-h^2[T_j,T_i]}+O(h^3)}U) - g(e^{ihT_i}e^{ihT_j}U)\Big) \\ &= \lim_{h\to0}\frac{1}{h^2}\Big(g(e^{ihT_i}e^{ihT_j}e^{-h^2[T_j,T_i]}U) - g(e^{ihT_i}e^{ihT_j}U)\Big) \\ &= \lim_{h\to0}\frac{1}{h^2}\Big(g(e^{-h^2[T_j,T_i]}U) - g(U)\Big) \\ &= \lim_{h\to0}\frac{1}{h^2}\Big(g(e^{ih^2 f_{ijk}T_k}U) - g(U)\Big) \\ &= f_{ijk}\partial_k g(U) \end{align}
alternative approach consider $\partial_i$ as a linear operator on the space of (sufficiently smooth) scalar functions on $G$. Then \begin{align} ([\partial_i,\partial_j]g)(U) &=(\partial_i\partial_j g)(U) - (i\leftrightarrow j) \\ &=\lim_h\frac{1}{h}\left((\partial_j g)(e^{ihT_i}U) - (\partial_j g)(U)\right) - (i\leftrightarrow j) \\ \end{align}
and now it should be \begin{align} (\partial_j g)(e^{ihT_i}U) &= \lim_h\frac{1}{h}\left(g(e^{ihT_j} e^{ihT_i}U) - g(e^{ihT_i})\right) \end{align}
which gets me to the same (presumably wrong) result.