Let $\textbf{S}(u,v):[0,1]^2 \rightarrow \mathbb{R}^3$ be a singular $2$-cube which is smooth. Note that $0 \leq u,v \leq 1$.
Let $B(\textbf{r})=B_x \ dy \wedge dz + B_y \ dz \wedge dx + B_z \ dx \wedge dy$ be a $2$-form on $\mathbb{R}^3$.
Show that $(\textbf{S}^*\textbf{B})(u,v)=\textbf{B}(\textbf{S}(u,v))\cdot \textbf{N}(u,v) \ du \wedge dv$,
where $\textbf{B}=(B_x, B_y, B_z)$ and $\displaystyle \textbf{N}=\frac{\partial \textbf{S}}{\partial u} \times \frac{\partial \textbf{S}}{\partial v}$
Please note that my (rough) definition of a pullback is $(F^*\beta)(x;u_1,\dots,u_n)=\beta(F(x); F'(x)u_1,\dots,F'(x)u_n)$.
$\begin{align}(\textbf{S}^*\textbf{B})(u,v) &= (\textbf{B} \circ \textbf{S})(u,v) \\ &= \textbf{B}(\textbf{S}(u,v)) \\ &= (B_x \ dy \wedge dz + B_y \ dz \wedge dx + B_z \ dx \wedge dy ) (\textbf{S}(u,v)) \\ &= (B_x \ dy \wedge dz)(\textbf{S}(u,v)) + (B_y \ dz \wedge dx)(\textbf{S}(u,v)) + (B_z \ dx \wedge dy) (\textbf{S}(u,v)) \\ &= B_x(\textbf{S}(u,v)) ( dy \wedge dz)(\textbf{S}(u,v)) + B_y(\textbf{S}(u,v)) ( dz \wedge dx)(\textbf{S}(u,v)) + B_z(\textbf{S}(u,v)) ( dx \wedge dy) (\textbf{S}(u,v)) \\ &= ... \end{align}$
Now just looking at the first term, I believe others with follow almost identically, I get that
$$\begin{align}B_x(\textbf{S}(u,v)) ( dy \wedge dz)(\textbf{S} (u,v)) &= B_x(\textbf{S}(u,v)) \left[ dy(\textbf{S} (u,v)) dz - dz(\textbf{S} (u,v)) dy)\right] \\ &= B_x(\textbf{S}(u,v)) \left[ dS_2 dz - dS_3 dy)\right] \\ &= B_x(\textbf{S}(u,v)) \left[\left( \frac{\partial S_2}{\partial u}du + \frac{\partial S_2}{\partial v}dv \right) dz - \left( \frac{\partial S_3}{\partial u}du + \frac{\partial S_3}{\partial v}dv \right) dy\right] \\ &= B_x(\textbf{S}(u,v)) \left[\left( \frac{\partial S_2}{\partial u}du + \frac{\partial S_2}{\partial v}dv \right) \left(\frac{\partial z}{\partial u}du+\frac{\partial z}{\partial v}dv\right) - \left( \frac{\partial S_3}{\partial u}du + \frac{\partial S_3}{\partial v}dv \right) \left(\frac{\partial y}{\partial u}du+\frac{\partial y}{\partial v}dv\right)\right] \\ \end{align}$$
Iam unsure whether this last equality is correct and how to proceed from here. I do not understand where the $du \wedge dv$ has gone or where it could come from.
Let $\omega=dx\wedge dy\wedge dz$ we have $B=i(\mathbf{B})\omega$. The simplest way to verify that $\mathbf{S}^*(B)=\mathbf{B}\cdot\mathbf{N}du\wedge dv$ is to apply both of sides to the basis vectors $\partial_u, \partial_v$. We have $$\mathbf{S}^*(B)(\partial_u, \partial_v)=B(\mathbf{S}_u, \mathbf{S}_v)$$ $$=i(\mathbf{B})\omega(\mathbf{S}_u, \mathbf{S}_v)=\omega(\mathbf{B}, \mathbf{S}_u, \mathbf{S}_v)$$ $$=\det(\mathbf{B},\mathbf{S}_u, \mathbf{S}_v)=\mathbf{B}\cdot(\mathbf{S}_u\times\mathbf{S}_v)$$ $$=\mathbf{B}\cdot\mathbf{N}$$
See Triple product for reference.