For reference this questions pertains to Adams and Essex, Calculus 2, chapter 17.4, Example 2.
From the book (slightly abbreviated):
A smooth (n-1)-manifold $\cal{M}$ in $\mathbb{R}^n$ is orientable if there exists a nonvanishing, normal vector field $\mathbf{n}(\mathbf{x})$ on $\cal{M}$. Since the tangent space $T_\mathbf{x}(\cal{M})$ to $\cal{M}$ has dimension (n-1), any n-1 linearly independent vectors $\mathbf{v}_1, \mathbf{v}_2, \cdots \mathbf{v}_{n-1}$ in $T_\mathbf{x}(\cal{M})$ will be perpendicular to $\mathbf{n}$ and the differential (n-1)-form
$\omega(\mathbf{x})(\mathbf{v}_1, \mathbf{v}_2, \cdots \mathbf{v}_{n-1}) = \begin{vmatrix} \vdots & \vdots & \cdots & \vdots \\ \mathbf{n}(\mathbf{x}) & \mathbf{v}_1 & \cdots & \mathbf{v}_{n-1} \\ \vdots & \vdots & \cdots & \vdots \end{vmatrix} $
orients $\cal{M}$.
Now, this expression for this differential (n-1)-form is a determinant of an $n \times n$ matrix. However, earlier in the book a differential (n-1) form is expressed as
$\omega(\mathbf{x})(\mathbf{v}_1, \mathbf{v}_2, \cdots \mathbf{v}_{n-1}) = \sum\limits_{1\leq i_1< i_2 < \cdots < i_{n-1} \leq (n-1)} a_{i_1 i_2 \cdots i_{n-1}} \begin{vmatrix} v_{1i_1} & v_{2i_1} & \cdots & v_{(n-1)i_1} \\ v_{1i_2} & v_{2i_2} & \cdots & v_{(n-1)i_2} \\ \vdots & \vdots & \ddots & \vdots \\ v_{1i_{n-1}} & v_{2i_{n-1}} & \cdots & v_{(n-1)i_{n-1}}\end{vmatrix} $
where the form is expressed as a sum over a determinant of an $(n-1) \times (n-1)$ matrix (as opposed to an $n \times n$ matrix above).
I cannot see how this sum of the $(n-1) \times (n-1)$ determinant can be transformed to the $n \times n$ determinant. I feel like it could have something to do with a cofactor expansion of the determinant but I just can't wrap my head around it. Any elucidation would be greatly appreciated.
OK, so I think I figured it out myself.
Looking at the case $k=n-1=2$ (2-form) in $\mathbb{R}^3$ (to make the construction clear), the value of a general differential 2-form can be written as
$ \omega(x)(v_1,v_2) = a_{12}(x) \begin{vmatrix} v_{11} & v_{21} \\ v_{12} & v_{22} \end{vmatrix} + a_{13}(x) \begin{vmatrix} v_{11} & v_{21} \\ v_{13} & v_{23} \end{vmatrix} + a_{23}(x) \begin{vmatrix} v_{12} & v_{22} \\ v_{22} & v_{23} \end{vmatrix} $
Here the first index on the $v_{ij}$ is the position of the vector in the argument and the second index is the component of the vector.
By cofactor expansion property of determinant this can be written as
$ \omega(x)(v_1,v_2) = \begin{vmatrix} a_{23} & v_{11}& v_{21} \\ a_{13} & v_{12} & v_{22} \\ a_{12} & v_{13}& v_{23} \end{vmatrix} $
Thus as long as $\mathbf{n}(\mathbf{x})$ is smooth and nonzero on the manifold we can associate
$ \mathbf{n}(\mathbf{x}) = \begin{bmatrix} a_{23}(x) \\ a_{13}(x) \\ a_{12}(x) \end{bmatrix} $
and the 2-form associated given by this components satisfies the requirement for orienting the manifold at a point $\mathbf{x}$.
This construction only works for (n-1)-manifolds in $\mathbb{R}^n$, that is the manifold has a 1-dimensional normal vector space.