Here $\phi^*(f_I)$ is defined by $$\phi^*(f_I)=f_I\circ\phi,$$ the composition of $\phi$ and $f_I$. This means $\phi^*(f_I)(\mathbf x)=f_I(\phi(\mathbf x))$; in other words, $\phi^*(f_I)$ is the function resulting from $f_I$ by substituting $y=\phi(\mathbf x)$. The pullback $\phi^*(\mathrm dy_I)$ is defined by replacing each $y_i$ with $\phi_i$. That is to say, if $I=(i_1,i_2,\dots,i_k)$ we put $$\phi^*(\mathrm dy_I)=\phi^*(\mathrm dy_{i_1}\mathrm dy_{i_2}\cdots\mathrm dy_{i_k})=\mathrm d\phi_{i_1}\mathrm d\phi_{i_2}\cdots\mathrm d\phi_{i_k}$$
But suppose $y = \phi(x)$, then I don't understand why
Define $\phi:U \to V$ where $U, V$ are open subsets of the Euclidean space $\mathbb{R}^n, \mathbb{R}^m$ respectively and $y$ be the coordinates on $V$ and $(x)$ be coordinates on $U$ and $y = \phi(x)$ be the coordinate change on the open subsets. My question is simply why when defining the pullback on $dy_I$, the pull back becomes $d\phi_I$? The expression $\phi^*(d\phi)$ doesn't make sense by itself.
$\phi^*(dy_I ) = \phi^*(d\phi) = d\phi_I$
Define $f_I: V \to R^n$ are the components of a k-form $\alpha \sum f_I dy_I$ on $V$