= Solution
A smooth <vector field> on a <smooth manifold> $M$ is a smooth <section of a vector bundle> $X:M\to TM$ of the <tangent bundle> projection $\pi$, so $\pi\circ X=\mathrm{id}_M$. In a coordinate chart it has the form $X=\sum_iX^i\partial_i$ with smooth component functions $X^i$.
Suppose a smooth <vector bundle isomorphism> $\Phi:M\times\mathbb R^n\to TM$ covering the identity is given. For the standard <basis> $e_1,\ldots,e_n$ of $\mathbb R^n$, put $X_i(p)=\Phi(p,e_i)$. These are smooth <vector fields>, and the <linear isomorphism> on each fibre makes $X_1(p),\ldots,X_n(p)$ a <basis> of $T_pM$.
Conversely, given such a global <frame of a vector bundle>, define
$$
\boxed{\Phi(p,a_1,\ldots,a_n)=\sum_{i=1}^na_iX_i(p).}
$$
It is smooth, covers the identity and is a <linear isomorphism> on each fibre. Its inverse is smooth as well: in any coordinate chart, the component columns of the $X_i$ form an invertible smooth <matrix> $C(p)$, and the fibre coordinates of the inverse are $C(p)^{-1}v$. <matrix> inversion is smooth on the invertible <matrices>. Thus \b[the <tangent bundle> is trivial exactly when a global frame exists], the defining property of a <parallelizable manifold>.
For a <Lie group> $G$, write $L_h(g)=hg$. A <left-invariant vector field> is a field satisfying
$$
X(hg)=(dL_h)_gX(g).
$$
Such a field is determined by $v=X(e)\in T_eG$, since $X(g)=(dL_g)_ev$. Conversely, this formula defines a left-invariant field, by the <chain rule> and $L_hL_g=L_{hg}$.
It also proves smoothness even if smoothness was not initially assumed. Choose a <smooth curve> $c(t)$ with $c(0)=e$, $c'(0)=v$. Then
$$
X(g)=\left.\frac{d}{dt}\right|_{t=0}g\,c(t).
$$
Multiplication is a <smooth map> on a <Lie group>, so differentiating its coordinate functions in the second variable yields a <smooth function> of $g$. This gives smooth components for $X$ in every local chart.
Choose a <basis> $v_1,\ldots,v_m$ of $T_eG$, where $m=\dim G$, and left translate it. Each $L_g$ is a <diffeomorphism>, with inverse $L_{g^{-1}}$, so $(dL_g)_e$ is a <linear isomorphism> and the translated fields are a <basis> at every $g$. The explicit <parallelization of a Lie group by left translations> is
$$
\boxed{G\times T_eG\longrightarrow TG,\qquad(g,v)\longmapsto(dL_g)_ev.}
$$
Its inverse sends $w\in T_gG$ to $(g,(dL_{g^{-1}})_gw)$. Identifying $T_eG$ with $\mathbb R^m$ gives the required product bundle. The differential facts used are linearity, the chain rule, $d(\mathrm{id})=\mathrm{id}$ and smooth dependence of the differential of a <smooth map> on its base point.
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