Now I read the book and realize that Lie derivative is introduced after the chapter on cohomology, if the order is reversed there is a very direct interpretation.
You want to prove:
If $f_0,f_1 : M\to N$ are smooth mapping which are homotopic, then
$$ f_0^* = f_1^* : H^k_{dR}(N)\to H^k_{dR}(M)$$
for all $k$.
Recall that the induced pullback mapping on $H^k$ is just $f_0^* [\alpha] = [f_0^*\alpha]$ and similar for $f_1$. So you need to show: for any $k$-form $\alpha$ on $N$, $ [f_0^*\alpha] =[ f_1^*\alpha]$, or $[f_1^*\alpha - f_0^*\alpha] = 0$.
That is, you want to write $f_1^*\alpha - f_0^*\alpha$ as $d$ of something. Note that by the fundamental theorem of calculus,
$$f_1^*\alpha - f_0^*\alpha = \int_0^1 \frac{\partial}{\partial t} (f_t^*\alpha) dt.$$
Here $f_t$ is the homotopy between $f_0$ and $f_1$. Of course it is not clear what the right hand side is. We want to give it a more intrinsic interpretation, so that we can check if the right hand side is really $d$ of something.
We let $F : M \times [0,1] \to N$ be the homotopy and $\iota_t : M\to M\times [0,1]$, $\iota_t (x) = (x, t)$ be the inclusion. Then $f_t = F\circ \iota_t$, thus $f_t^* = \iota_t^* \circ F^*$ and
$$\begin{split}
\frac{\partial}{\partial t} (f_t^* \alpha) (x)&= \frac{\partial }{\partial t} (\iota_t^* (F^*\alpha) (x)) \\
&= \frac{\partial}{\partial t} (F^*\alpha)(x, t)\\
&= \mathscr{L}_T (F^*\alpha),
\end{split}$$
$\mathscr L_T$ is the Lie derivative along the vector $T:=\frac{\partial}{\partial t}$ (as a vector field on $M\times [0,1]$). Now the Cartan's magic formula gives (for any differential form $\omega$, vector fields $X$)
$$ \mathscr L_X \omega = \iota_X d\omega + d \iota_X \omega.$$
So we have
$$\begin{split}
\int_0^1 \frac{\partial}{\partial t} (f_t^*\alpha) dt &= \int_0^1 \mathscr L_T (F^*\alpha) dt \\
&= \int_0^1 \big( \iota_T d(F^*\alpha) + d\iota_T (F^*\alpha)\big) dt \\
&= \int_0^1 \iota_T F^* (d\alpha) dt+ d \left( \int_0^1 \iota_T (F^*\alpha) dt\right)
\end{split}$$
Note that the integration are exactly the homotopy operator $h$ constructed: so
$$ \int_0^1 \frac{\partial}{\partial t} (f_t^*\alpha) dt = h(d\alpha) + d(h(\alpha)).$$
So we have the next best thing: the right hand side in general is not $d$ of something, but it is when $\alpha$ is closed. This proves the theorem.
Of course I am just hiding everything in the Cartan's magic formula. The formula is commonly proved by direct calculation. A more fancy/geometric argument is suggested in Arnold's classical mechanics here. Note that the latter one also use a homotopy operator.
Best Answer
This is true for any two $f,g$. You do not need the fact that they are homotopic.
Let $M' = M + M$ be the disjoint union of two copies of $M$. Then $f, g$ induces a smooth map $\phi : M' \to N$ by taking $f$ on the first summand and $g$ on the second.
The set of regular values of $\phi$ is non-empty (it is dense in $N$). Any regular value of $\phi$ is a common regular value of $f,g$.