With regards to your specific query, the local trivialisations commute with the fibrewise linear structure on $E$. So if $\varphi:p^{-1}(U)\xrightarrow{\cong}U\times \mathbb{R}^n$ is such a local trivialisation then for each $b\in U$ we have
$$\varphi\left(\sum_{i=1}^n t_i\cdot s_i(b)\right)=\sum_{i=1}^n t_i\cdot \varphi (s_i(b))$$
where the right-hand side is interpreted as the sum in $\mathbb{R}^n\cong\{b\}\times\mathbb{R}^n$. The point is that the sections $s_i$ are continuous with respect to the variable $b\in U$ so if for each $i=1,\dots,n$ we let $\hat s_i:U\rightarrow \mathbb{R}^n$ be the composite
$$\hat s_i:U\xrightarrow{s_i} E_U\xrightarrow{\varphi}U\times \mathbb{R}^n\xrightarrow{pr_2} \mathbb{R}^n$$
then we get a family of continuous $\mathbb{R}^n$-valued maps on $U$, and our first equation tells us that the composite $\varphi\circ h|_{U\times\mathbb{R}^n}:U\times\mathbb{R}^n\rightarrow U\times \mathbb{R}^n$ is equal to the composite
$$U\times \mathbb{R}^n\xrightarrow{\Delta\times 1}U^{n+1}\times\mathbb{R}^n\xrightarrow{shuf}U\times (\mathbb{R}\times U)^n\xrightarrow{1\times\prod(1\times\hat s_i)}U\times (\mathbb{R}\times \mathbb{R}^n)^n\xrightarrow{1\times m^n}U\times (\mathbb{R}^n)^n\xrightarrow{1\times\oplus}U\times\mathbb{R}^n$$
where $\Delta:U\rightarrow U^{n+1}$ is the $(n+1)$-fold diagonal, the second map $shuf$ shuffles the coordinates appropriately, $m:\mathbb{R}\times\mathbb{R}^n\rightarrow \mathbb{R}^n$ is scalar multiplication and $\oplus:\mathbb{R}^n\times\dots\times\mathbb{R^n}\rightarrow\mathbb{R}^n$ $(x_1,\dots,x_n)\mapsto x_1+\dots+x_n$ is the iterated vector addition in $\mathbb{R}^n$.
It should be clear from this presentation that $\varphi\circ h$ is continuous (and even smooth if you work in the $C^\infty$ category).
Suppose $M$ is embedded into the standard inner product space $\Bbb{R}^n$.
Since $TM \oplus NM \cong M \times \Bbb{R}^n$, by the Whitney sum formula, $w(TM) \cdot w(NM)=w(TM \oplus NM)=w(M \times \Bbb{R}^n)=1 \in H^0(M;\Bbb{Z}_2)$ where $w$ denotes the total Stiefel-Whitney class.
Specifically, $w_1(TM) + w_1(NM)=w_1(M \times \Bbb{R}^n)=0 \in H^1(M;\Bbb{Z}_2)$, which means $w_1(NM) = w_1(TM) \ne 0$.
Best Answer
If $w_i(E) \neq 0$, then $E$ cannot admit $n-i+1$ linearly independent sections, but the converse is not true. That is, if $E$ does not admit $n - i + 1$ linearly independent sections, we could still have $w_i(E) = 0$. For example, the bundle $TS^2 \to S^2$ does not admit a linearly independent section (i.e. a nowhere-zero section) by the Poincaré-Hopf Theorem. However, we have $w_2(TS^2) = 0$; note that $1 = n - i + 1$ for $i = 2$.
In particular, if a rank $n$ bundle $E$ does not admit $n$ linearly independent sections, we could still have $w_1(E) = 0$, i.e. $E$ need not be non-orientable.