Let $M$ be a smooth manifold, $X$ a smooth outward pointing vector field on $\partial M$, and $\omega$ an orientation form on $M$. In Tu's Introduction to Manifolds, he proves that the contraction of $\omega$ with $X$ is nowhere vanishing on $\partial M$. In this proof he says,
Let $e_1, \ldots, e_{n-1}$ be a basis for $T_p(\partial M)$. Then $X_p, e_1, \ldots, e_{n-1}$ is a basis for $T_pM$.
How do we know that this is true? What ensures $X_p$ to be linearly independent from the other $e_i$?
$X_p$ being outward pointing implies in particular that $X_p \in T_pM \setminus T_p(\partial M)$. Since $\{e_1,\ldots,e_{n-1}\}$ spans $T_p(\partial M)$, $X_p$ is not a linear combination of these latter vectors.