Let $$ F:\Bbb R^n\times[0,1]\to\Bbb R^n $$ smooth; let $(s,u)$ be the coordinates on $\Bbb R^n\times[0,1]$ and let $\omega(\zeta)=f(\zeta)d\zeta$ be an $n$-form.
The pull back is defined as $$ F^*\omega(s,u)=f(F(s,u))\det\left(\frac{\partial F}{\partial s}(s,u)\right)ds $$ which is again an $n$-form on $\Bbb R^n\times[0,1]$.
According to Spivak (exercise 22 pag 236, chapt.7), we can write uniquely every such form as $$ du\wedge\alpha(s,u)+\beta(s,u) $$ where $\alpha$ is an $n-1$ form and $\beta$ is a $n$-form. Given that, we can write $$ \alpha(s,u)=\sum_{j=1}^nf_j(s,u)ds[j] $$ where $ds[j]=ds_1\wedge\cdots\wedge\widehat {ds_j}\wedge\cdots\wedge ds_n$ (the hat denotes omission) and $\beta(s,u)=f_0(s,u)ds$ for some suitable $f_0$.
I read that $$ f_j(s,u)=f(F(s,u))\det\left(\frac{\partial F}{\partial u},\frac{\partial F}{\partial s_1},\cdots,\widehat{\frac{\partial F}{\partial s_j}},\cdots,\frac{\partial F}{\partial s_n}\right) $$ I'm trying to figure out why does $\alpha$ has to be of that form, but I can't. Any hint would be appreciated.