= Solution
A smooth <vector field> on a <smooth manifold> $M$ is a smooth section $X:M\to TM$ of its <tangent bundle>. It acts on smooth functions by $Xh(p)=dh_p(X_p)$. In a <manifold chart>, $X=\sum_iX^i\partial_i$. The <Lie bracket of vector fields> is the commutator of their actions on functions:
$$
[X,Y]h=X(Yh)-Y(Xh).
$$
Expanding in coordinates makes all second derivatives cancel and gives
$$
[X,Y]=\sum_i\left(\sum_jX^j\partial_jY^i-Y^j\partial_jX^i\right)\partial_i.
$$
Thus the commutator is again a smooth <vector field>, not a second-order operator. The coordinate expression represents the intrinsic derivation on functions, so agrees on overlaps.
For a <diffeomorphism> $f:M\to N$, its <pushforward of a vector field> is
$$
(f_*X)_q=df_{f^{-1}(q)}X_{f^{-1}(q)}.
$$
For every smooth function $h$ on $N$, the <chain rule> gives $((f_*X)h)\circ f=X(h\circ f)$. Applying this twice yields
$$
([f_*X,f_*Y]h)\circ f=X(Y(h\circ f))-Y(X(h\circ f))=([X,Y](h\circ f))=((f_*[X,Y])h)\circ f.
$$
Smooth functions determine a tangent vector, and $f$ is surjective, so
$$
\boxed{f_*[X,Y]=[f_*X,f_*Y].}
$$
On a <Lie group>, a <left-invariant vector field> satisfies $(L_g)_*X=X$ for every left translation $L_g(h)=gh$. Given $\xi\in T_eG$, set
$$
X_\xi(g)=(dL_g)_e\xi.
$$
This field is smooth by smoothness of multiplication. The equality $L_aL_g=L_{ag}$ proves left invariance. Conversely, left invariance forces $X(g)=(dL_g)_eX(e)$, so $X_\xi$ is the unique left-invariant field with value $\xi$ at $e$. This is a linear isomorphism between $T_eG$ and the space of <left-invariant vector fields>.
The pushforward identity applied to $L_g$ shows that $[X_\xi,X_\eta]$ is again left invariant. Consequently
$$
[\xi,\eta]=[X_\xi,X_\eta](e)
$$
defines a <Lie bracket> on $T_eG$. Bilinearity and antisymmetry follow from the operator commutator. Its <Jacobi identity> follows by expanding the three nested commutators: the six triple compositions cancel in pairs. These properties transfer through the linear isomorphism, making $T_eG$ the <Lie algebra> of $G$.
For the <general linear group> $G=\mathrm{GL}(n,\mathbb R)$, use matrix-entry coordinates $x_{pq}$. It is an open subset of $M_n(\mathbb R)$, so every tangent space is naturally $M_n(\mathbb R)$. Since $L_g(h)=gh$, its differential sends $A$ to $gA$, and hence
$$
X_A=\sum_{p,q}(gA)_{pq}\frac{\partial}{\partial x_{pq}},\qquad X_A(x_{pq})=\sum_kx_{pk}A_{kq}.
$$
For $A,B\in M_n(\mathbb R)$,
$$
X_A(X_B(x_{pq}))=\sum_k(gA)_{pk}B_{kq}=(gAB)_{pq},
$$
and reversing $A,B$ gives $(gBA)_{pq}$. Thus $[X_A,X_B]=X_{AB-BA}$; evaluating at the identity gives the <matrix commutator>
$$
\boxed{[A,B]=AB-BA.}
$$
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