Let $G$ be a Lie group. Its Lie algebra, when considered as a vector space, is given by $T_e G$, the tangent space of $G$ at the identity element $e$.
However, in order to fully describe the Lie algebra associated with $G$ we must also describe the commutator $\left[\cdot,\,\cdot\right]:T_eG^2\to T_eG$ and show that it obeys the three axions: bilinearity, alternativity and the Jacobi identity.
Following Wikipedia, one defines first for any $g\in G$ left multiplication on $G$ by $g$ as $$ l_g:G\to G$$ given by $$ x\mapsto gx$$and a left-multiplication invariant vector field $X\in TG$ is one which obeys for every $\left(g,\,h\right)\in G^2$ $$ \left(d l_g\right)_h X_h = X_{gh} $$ where $\left(d l_g\right)_h:T_hG\to T_{gh}G$ is the differential of $l_g$ at $h$. We define $Lie\left(G\right)$ as the vector space of left-multiplication invariant vector fields.
Apparently $G\times Lie\left(G\right)$ is isomorphic to $TG$ via $$\left(g,X\right)\mapsto\left(g,X_g\right)$$ so that we get a $\natural$ isomorphism $Lie\left(G\right)$ with $T_eG$ given by $$Lie\left(G\right)\ni X \mapsto X_e\in T_eG$$This last isomorphism we call $\phi$.
Now apparently there is a Lie bracket structure on $TG$ which induces a Lie bracket structure on $Lie\left(G\right)$: for any $X$ and $Y$ in $TG$ we have $$ \left[X,\,Y\right]_{TG}= X\left(Y\left(\cdot\right)\right)-Y\left(X\left(\cdot\right)\right)$$ This in turn can be pushed forward to a define our desired Lie bracket on $T_e G \equiv \frak{g} $ via: If $X$ and $Y$ are any tangent vectors in $T_eG$ then $$\left[X,\, Y\right]_{T_eG}=\phi\left[\phi^{-1}\left(X\right),\,\phi^{-1}\left(Y\right)\right]_{TG}$$
My question is: is there now any easier way to describe $\left[\cdot,\,\cdot\right]:T_eG^2\to T_eG$? I tried for example to write it in coordinates and it was a mess. How do you "easily" see how $\left[X,\, Y\right]_{T_eG}$ acts on any scalar function $f$ given some $X$ and $Y$ in $T_eG$?