Local frames and local trivializations are equivalent data

If $(s_1,\dots ,s_r)$ is a local frame on $U$, then [L1] gives for each $p\in U$ and $v\in E_p$ unique scalars $a_1,\dots ,a_r$ with $v={\sum }_i{a}_i{s}_i(p)$. Around any $p\in U$, choose an existing vector bundle chart $\mathrm{\Psi }:E{\mathrm{\mid }}_W\to W\times {\mathbb{R}}^r$ with $W\subseteq U$. Writing $\mathrm{\Psi }(s_i(x))=(x,a_i(x))$, the column vectors $a_i(x)$ form a smooth matrix $A(x)\in GL(r,\mathbb{R})$ because the $s_i(x)$ are a basis of $E_x$. Thus the local coordinate map is ${\mathrm{\Phi }}_W(v)=(x,A(x{)}^{-1}u)$ when $\mathrm{\Psi }(v)=(x,u)$, so ${\mathrm{\Phi }}_W$ and its inverse $(x,\lambda )\mapsto {\mathrm{\Psi }}^{-1}(x,A(x)\lambda )$ are smooth. By uniqueness of the coordinates from [L1], these local formulas agree on overlaps and patch to a vector bundle chart $\mathrm{\Phi }:E{\mathrm{\mid }}_U\to U\times {\mathbb{R}}^r$.

Conversely, if $\mathrm{\Phi }:E{\mathrm{\mid }}_U\to U\times {\mathbb{R}}^r$ is a vector bundle chart, let $e_1,\dots ,e_r$ be the standard basis of ${\mathbb{R}}^r$ and set $s_i(p)={\mathrm{\Phi }}^{-1}(p,e_i)$. Then each $s_i$ is a smooth local section, and the vectors $s_i(p)$ form a basis of $E_p$ because $\mathrm{\Phi }{\mathrm{\mid }}_{E_p}$ is a linear isomorphism.

Applying the second construction to the chart from step 1.1 recovers the original frame because $\mathrm{\Phi }(s_i(p))=(p,e_i)$ by construction. Applying the first construction to the sections from step 1.2 recovers the original chart because the resulting coordinates are exactly the fibre coordinates already read by $\mathrm{\Phi }$. Hence local frames and local trivializations are inverse constructions.