Let $\omega=\sum_j \omega_jdy^j$. We want to show that $d(F^{*}\omega)=F^{*}(d\omega)$, where $F: M \to N$ is a map from manifold $M$ to manifold $N$. Local coordinate systems of $M, N$ are $(y^1, \ldots, y^m), (x^1, \ldots, x^n)$ respectively.
Now
$F^{*}\omega=\sum_{j,k}\omega_j\frac{\partial y^j}{\partial x^{k}}dx^k,$
$d(F^{*}\omega)=\sum_{j,k,i}\frac{\partial ( \omega_j\frac{\partial y^j}{\partial x^k} )}{\partial x^i}dx^i \wedge dx^k = \sum_{j,k,i}\frac{\partial \omega_j}{\partial x^i} \frac{\partial y^j}{\partial x^k} dx^i \wedge dx^k + \sum_{j,k,i} \omega_j \frac{\partial^2 y^j}{\partial x^i \partial x^k} dx^i \wedge dx^k.$
$d\omega=\sum_{j,l}\frac{\partial \omega_j}{\partial y^i} dy^l \wedge dy^j,\qquad F^{*}d\omega = \sum_{i,j,k,l} \frac{\partial \omega_j}{\partial y^l}\frac{\partial y^l}{\partial x^i} \frac{\partial y^l}{\partial x^i} dx^i \wedge dx^k = \sum_{i,j,k} \frac{\partial \omega_j}{\partial x^i} \frac{\partial y^l}{\partial x^i} dx^i \wedge dx^k.$
My question is why $\displaystyle\sum_{j,k,i} \omega_j \frac{\partial^2 y^j}{\partial x^i\; \partial x^k} dx^i \wedge dx^k = 0$? Thank you very much.