If $M,N$ are smooth manifolds and $F: M \to N$ is a surjective smooth submersion. A tangent vector $v \in T_p M $ is called vertical if $d F_p (v) = 0$. Now suppose $\omega \in \Omega^k (M)$, I want to show that if $v \lrcorner \omega _p =0 $ and $v \lrcorner d \omega _p = 0$ for all $p \in M$ and $v \in T_p M$ vertical, then there exists a $\eta \in \Omega ^k (N)$ such that $\omega = F^* \eta $.
I have no idea how to get started. Any help is appreciated.
You know that we can choose local coordinates on $M$ and $N$ so that $F(x^1,\dots,x^n,x^{n+1},\dots, x^m)=(x^1,\dots,x^n)$. Write $$\omega = \sum_I a_I(x)dx^I + \sum_{I',J} a_{I'J}dx^J\wedge dx^{I'},$$ where $I$ is a $k$-multiindex involving only $1,\dots,n$ and $J$ is a $j$-multiindex involving only $n+1,\dots,m$ for various $j=1,\dots,k$, and the length of $I'$ is $k-j$. Because $\omega$ annihilates vertical vectors, the terms in the latter sum are all $0$. Now we wish to show that $a_I(x) = a_I(x^1,\dots,x^n)$, i.e., the coefficients of the first terms don't vary along the fibers. Do this by differentiating and using the hypothesis that $d\omega$ annihilates vertical vectors.
Incidentally, I think you need the fibers of $F$ to be connected. Can you see why?
(EDIT) Remark: You can do this without local coordinates, as well. Define $$\eta_q(w_1,\dots,w_k) = \omega_p(v_1,\dots,v_k)$$ for any $p$ with $F(p)=q$ and for any vectors $v_i$ with $dF_p(v_i)=w_i$. You can check that because $\omega$ annihilates vertical vectors, this definition doesn't depend on the choice of $v_i$. And, using the Cartan formula for Lie derivative and the fact that $d\omega$ annihilates vertical vectors, you can check that the definition doesn't depend on the choice of $p$.