Let M be a smooth manifold, $i_t: M \rightarrow M \times [0,1]$ such that $i_t(x) = (x,t)$. Let $\omega \in \Omega^p(M \times [0,1])$ such that $\omega = f(x,t)dt \wedge dx^I$, where $I$ is a multi-index and $f$ a smooth function.
I need to prove that $i^*_t (\omega)=0$, but I have some troubles.
My attempt is, given $X_1,\dots,X_{p} \in T_x M$: $$i^*_t (\omega)(X_1,\dots,X_{p}) = i^*_t(f(x,t)dt \wedge dx^I) = i^*_t(f(x,t)dt) \wedge i^*_t(f(x,t)dx^I)= f(x,t)|_{(x,t)}\big(\frac{\partial i_t}{\partial x}(X_1),\dots,\frac{\partial i_t}{\partial x}(X_p)\big)$$
and I don't know how to proceed. Can anyone help me with a direct computation (if my attempt is right)?
First of all, you mistakenly have two "copies" of $f(x,t)$. $2\,dx\wedge dy$ is not equal to $(2\,dx)\wedge (2\,dy)$.
Second, where did the $i_t^*dt$ term go? If you restrict $dt$ to a submanifold where $t$ is constant, what do you get? (The notation is perhaps too confusing. I would write $i_s(x) = (x,s)$ and not repeat the variable.)