If $M^n$ is a submanifold of $\mathbb{R}^{n+k}$ such there are L.I. vector fields $v_1,\ldots,v_k$ such that $v_i \perp T_pM$ for every $p \in M$ then $M$ is orientable.
My attempt is:
Once $M^n$ is embedded on $\mathbb{R}^{n+k}$ take $dx_1\wedge \ldots dx_n \wedge dx_{n+1} \wedge \ldots \wedge dx_{n+k}$ the standard volume form on $\mathbb{R}^{n+k}.$
Lets define
$$\mu_M = \iota_{v_1,\ldots,v_k}dx_1 \wedge dx_2 \wedge \ldots \wedge dx_n\wedge \ldots\wedge dx_{n+k}.$$
This is a volume form to $M$!
We say that $M$ is positively oriented if $\mu_M (X_1,\ldots,X_n) > 0$ and negatively otherwise.
Is this right?
Yes, it's correct. (You should probably comment that $\mu_M$ is nowhere-zero because of the linear independence of the $v_i$. Can you check that?) You also have a trivialization of the normal bundle, so it's orientable, and that means the tangent bundle is orientable as well. (Any time two of three vector bundles in a short exact sequence are orientable, so is the remaining one.)