I've found similar questions, but their accepted answers either use forms or don't quite answer what I'm looking for.
To make clear the definitions I'm allowed to use, an orientation on an $m$-manifold $M$ is a choice of an orientation on each tangent space $T_pM$, i.e. an equivalence class of an ordered basis $(v_1, \ldots, v_m)$ on $T_pM$ so that the determinant of the change of basis to any other basis in the equivalence class is positive, and if $(U, \phi)$ is chart containing $p \in M$ and $d_p\phi:T_pM \rightarrow \mathbb{R}^m$, then the ordered basis $(d_p\phi(v_1) ,\ldots, d_p\phi(v_n))$ is equivalent to the standard basis on $\mathbb{R}^m$.
Let $M$ be an oriented manifold with boundary. Let $p \in \partial M$ and $(v_1, \ldots v_m)$ be an ordered basis on $T_pM$ that represents the given orientation on $M$ such that $v_1$ is outwards-pointing, i.e. the $m$-th coordinate of $d_p\phi(v_1) \in \mathbb{R}^m$ is negative, and $v_2, \ldots, v_m$ are in $T_p(\partial M) \subset T_pM$. Define the induced orientation on $\partial M$ to be equivalence class determined by $(v_2, \ldots v_m)$.
I need to show that this induced orientation is in fact an orientation on $\partial M$ according to the definition I've given. I think I'm supposed to prove that $(d_p\phi(v_2) ,\ldots, d_p\phi(v_n))$ is equivalent to the "standard basis on $\mathbb{R}^{m-1}$". I use quotes because these vectors lie in $\mathbb{R}^m$, so I would need to look at an $m-1$ dimensional subspace. However, I'm not sure which subspace to look at and which "direction" to look at the subspace.
For example, in $\mathbb{R}^3$ with basis $(e_1, e_2, e_3)$, I know that $(e_2, e_3)$ is a basis for a $2$-dimensional subspace of $\mathbb{R}^3$. But looking at this subspace from different sides changes the orientation of this basis.
I'm also confused about the significance of $v_1$ being an outwards pointing vector. Maybe resolving this confusion would help resolve the previous confusion.