Edit: I am inclined to believe that the overall answer is going to be: it doesn't matter which sense we call positively oriented as long as we are consistent during our traversal.
The following is from C.H. Edwards's Advanced Calculus of Several Variables, Section V6 Stokes' Theorem. I'm pretty sure there are some typos in the indexing. The quoted material appears contiguously in the book. I have interleaved my text.
So we start with the unit cube \begin{align*} \mathcal{I}^{k}= & \left\{ \left(x_{1},\dots,x_{k}\right)\in\mathbb{R}^{k}:\text{each }x_{i}\in\left[0,1\right]\right\} \end{align*} in $\mathbb{R}^{k}.$ Its boundary $\partial\mathcal{I}^{k}$ is the set of all those points $\left(x_{1},\dots,x_{k}\right)\in\mathcal{I}^{k}$ for which one of the $k$ coordinates is either $0$ or $1;$ that is, $\partial\mathcal{I}^{k}$ is the union of the $2^{k}$ different $\left(k-1\right)$-dimensional faces of $\mathcal{I}^{k}.$ These faces are the sets defined by \begin{align*} \mathcal{I}_{i,\epsilon}^{k-1}= & \left\{ \left(x_{1},\dots,x_{k}\right)\in\mathcal{I}^{k}:x_{i}=\epsilon\right\} \end{align*} for each $i=1,...,k$ and $\epsilon=0$ or $\epsilon=1$ (see Fig. 5.32). The $\left(i,\epsilon\right)^\text{th}$ face $\mathcal{I}_{i,\epsilon}^{k-1}$ of $\mathcal{I}^{k}$ is the image of the unit cube $\mathcal{I}^{k-1}\subset\mathbb{R}^{k-1}$ under the mapping $\iota_{i,\epsilon}:\mathcal{I}^{k-1}\to\mathbb{R}^{k}$ defined by \begin{align*} \iota_{i,\epsilon}\left(x_{1},\dots,x_{k-1}\right) & =\left(x_{1},\dots,x_{i-1},\epsilon,x_{i},\dots,x_{k-1}\right). \end{align*}
Is this correct? Or should it be
\begin{align*} \iota_{i,\epsilon}\left(x_{1},\dots,x_{k-1}\right) & =\left(x_{1},\dots,x_{i-1},\epsilon,x_{i+1},\dots,x_{k}\right)? \end{align*}
Edit: the above was answered in favor of Edwards in a comment. My curiosity about the mapping had to do with the number of permutation needed to put the inserted dimension at the end.
The mapping $\iota_{i,\epsilon},$ serves as an orientation for the face $\mathcal{I}_{i,\epsilon}^{k-1}.$ The orientations which the faces (edges) of $\mathcal{I}^{2}$ receive from these mappings are indicated by the arrows in Fig. 5.32. We see that the positive (counterclockwise) orientation of $\partial\mathcal{I}^{2}$ is given by \begin{align*} \partial\mathcal{I}^{2}= & -\mathcal{I}_{1,0}^{1}+\mathcal{I}_{1,1}^{1}-\mathcal{I}_{2,1}^{2}+\mathcal{I}_{2,0}^{2}. \end{align*}
I'm pretty sure all the superscripts on the RHS of the above equation should be $1$, indicating a $1$-dimensional "face"; that is \begin{align*} \partial\mathcal{I}^{2}= & -\mathcal{I}_{1,0}^{1}+\mathcal{I}_{1,1}^{1}-\mathcal{I}_{2,1}^{1}+\mathcal{I}_{2,0}^{1}. \end{align*} Is that correct?
If $\omega$ is a continuous 1-form, it follows that the integral of $\omega$ over the oriented piecewise $\mathscr{C}^{1}$ closed curve $\partial\mathcal{I}^{2}$ is given by \begin{align*} \int_{\partial\mathcal{I}^{2}}= & -\int_{\iota_{1,0}}\omega+\int_{\iota_{1,1}}\omega-\int_{\iota_{2,1}}\omega+\int_{\iota_{2,0}}\omega\\ = & \sum_{i=1}^{2}\sum_{\epsilon=0,1}\left(-1\right)^{i+\epsilon}\int_{\iota_{i,\epsilon}}\omega. \end{align*}
Evidently, this integral is over the boundary of the unit square in the $x_1\times x_2$ coordinate plane. That is, $\partial\mathcal{I}_{3,0}^{2}.$ Or, perhaps we could think of it as $\partial\mathcal{I}_{3,\varepsilon}^{2},$ where $\varepsilon=0\vert 1$ indicates the 2D face whose boundary we are integrating over. So I ask: What is the domain of integration in the above equation?
The integral of a $\left(k-1\right)$-form $\alpha$ over $\partial\mathcal{I}^{k}$ is defined by analogy. That is, we define \begin{align*} \int_{\partial\mathcal{I}^{k-1}}\alpha= & \sum_{i=1}^{k}\sum_{\epsilon=0,1}\left(-1\right)^{i+\epsilon}\int_{\iota_{i,\epsilon}}\alpha. \end{align*} As in Section 5, the integral over $\mathcal{I}^{k}$ of the $k$-form $f\mathrm{d}x_{1}\wedge\dots\wedge\mathrm{d}x_{k}$ is defined by \begin{align*} \int_{\mathcal{I}^{k}}f\mathrm{d}x_{1}\wedge\dots\wedge\mathrm{d}x_{k}= & \int_{\mathcal{I}^{k}}f. \end{align*}
Figure 5.32
My graphics
These were produced a long time ago when working with the stress tensor. I relabeled the "blowup" image in an effort to match Edwards.
$\color{red}{\text{front:}x=1} ,\color{Green}{\text{ right:}y=1} , \color{blue}{\text{ top:}z=1}, $
$\color{cyan}{\text{back:}x=0} ,\color{magenta}{\text{ left:}y=0} , \bbox[black]{\color{yellow}{\text{ bottom:}z=0}}$
Original blowup diagram using coordinates $x,y,z$
Relabeled blowup diagram to match Edwards
As drawn, the positive sense of traversal is around the outward normal. Should the faces with $\epsilon=0$ be traversed from the inside (right thumb pointing in the positive coordinate direction), or from the outside (right thumb pointing in the direction of the outward normal)?
Edit: This may actually constitute an answer.
The following modified depiction seems to move closer to understanding. If we cheat, and treat the wedge product as a cross product, we find that each surface has an outward pointing normal. So in the case of those faces I have designated as $'+',$ the normals are the corresponding basis vectors. In the $'-'$ cases, the outward normals are the negatives of the basis vectors.
