Suppose $M$ and $N$ are smooth manifolds of dimensions $m$ and $n$ respectively. Suppose we have a smooth map $F:M\times [0,1]\to N$. Clearly, for $t\in[0,1]$, $F$ defines a smooth map $F_t:M\to N$. Now suppose that $\alpha\in\Omega^k(N)$ is a differential $k$-form. The claim is that
$$F^*\alpha=F_t^*\alpha+dt\wedge\gamma,$$
where $\gamma=\partial/\partial t \lrcorner F^*(\alpha)$, ie $\gamma$ is the contraction of the form $F^*\alpha$ to the vector $\partial/\partial t.$
The result makes sense but I don't see how I can prove it from the definition of a pullback directly. The definition I know is the pointwise pullback, ie for $(p,t)\in M\times[0,1]$ and $v^1_{(p,t)},\ldots,v^k_{(p,t)}\in T_{(p,t)}(M\times[0,1])$ are $k$ linearly independent tangent vectors at $(p,t)$,
we have:
$$F^*_{(p,t)}\left(\alpha_{F(p,t)}\right)\left(v^1_{(p,t)},\ldots,v^k_{(p,t)}\right)=\alpha_{F(p,t)}\left(F_*\left(v^1_{F(p,t)}\right),\ldots,F_*\left(v^k_{(p,t)}\right)\right)$$
where $F_*=F^*_{(p,t)}:T_{(p,t)}(M\times [0,1])\to T_{F(p,t)}N$ is the push-forward map.
Note: This formula is a main ingredient in the proof of the Point-Caré Lemma which states that $H^k(\mathbb{R}^m)$=0 for $k>0$.
Best Answer
Locally, let's think about the case $F\colon U\times [0,1]\to\Bbb R^m$, where $U\subset\Bbb R^n$ is open. Let $(x^1,\dots,x^n)$ be coordinates on $U$ and $(y^1,\dots,y^m)$ be coordinates on $\Bbb R^m$. Then $$F^*dy^\alpha = \sum_i f^\alpha_i(x,t)\,dx^i + g^\alpha(x,t)\,dt$$ for $1\le\alpha\le m$. What happens when you consider $F^*(dy^\alpha\wedge dy^\beta)$? (You can then generalize to the case of $k>2$ without writing unwieldy formulas.) You can check that you get \begin{multline*} F^*(dy^\alpha\wedge dy^\beta) = F^*dy^\alpha\wedge F^*dy^\beta \\ = \sum_{i,j} f^\alpha_i(x,t)f^\beta_j(x,t)\,dx^i\wedge dx^j + \\ dt\wedge\sum_i \big(g^\alpha(x,t)f^\beta_i(x,t)- g^\beta(x,t)f^\alpha_i(x,t)\big)dx^i. \end{multline*} Do you see that this fits the form you want?