Exercise 6.50 in Bott Tu Algebraic Topology The problem asks to show that
$$H^k(f^*) \simeq H^k(g^*)$$
For homotpic maps $f$ and $g$.
Where, given a map $f:M\to S$, $H^k(f^*)$ is defined by first defining the complex $\Omega^k(f^*) = \Omega^k(M)\oplus\Omega^{k-1}(S)$ and then the differential operator $d_f:\Omega^k(f^*)\to\Omega^{k+1}(f^*)$ by
$$d_f(\omega,\theta) = (d\omega,f^*\omega-d\theta)$$
Since $\Omega^k(f^*) = \Omega^k(g^*) = \Omega^k(M)\oplus\Omega^{k-1}(S)$, it seems to me that the proof should involve showing that $\ker(d_f) = \ker(d_g)$ and $\text{Im}(d_f) = \text{Im}(d_g)$.
Working from Corollory 4.1.2 (homotopic maps induce the same maps in cohomology) gives
$$f^*\omega = g^*\omega + d\lambda$$
for any closed form $\omega\in\Omega^k(M)$ ($\lambda\in\Omega^{k-1}(S)$). From here, I can show the images are the same, since
$$d_f(\omega,\theta) = (d\omega,f^*\omega-d\theta) = (d\omega,g^*\omega-d(\theta-\lambda)) = d_g(\omega,\theta-\lambda)$$
and so any form in the image of $d_f$ is also in the image of $d_g$ and vice versa, but I cannot seem to get a similar result on the kernels, instead, all I get is that if $d_f(\omega,\theta) = 0$, then $d_g(\omega,\theta) = (0,-d\lambda).$
 A: (ended up posting this as an answer since it turned out so long)
I think I managed to do it. Going back to the proof of the Poincaré lemmas in the book, they show that the composition of the $0\text{-section}$ $s_0$ with the projection $\pi$ on $M\times\mathbb{R}$ is chain homotopic to the identity
$$\mathbb{1} - \pi^*\circ s_0^* = (-1)^{q-1}(dK - Kd)$$
where they define $K$ as integration of over fibres. First, one can extend this to any constant $r\in\mathbb{R}$ section $s_r$ by altering the definition of $K$ so that its action on type II forms from
$$(\pi^*\phi)f(x,t)dt\mapsto(\pi^*\phi)\int_0^t f$$
to
$$(\pi^*\phi)f(x,t)dt\mapsto(\pi^*\phi)\int_r^t f$$
This then allows us to write
$$\mathbb{1}-\pi^*\circ s_r^* = (-1)^{q-1}(dK_r - K_rd)$$
With $K_r$ following the new definition.
From here, we note that
\begin{align}
\pi^*\circ(s_0^*-s_1^*) &= \pi^*\circ s_0^* - \pi^*\circ s_1^*\\
& = -(\mathbb{1}-\pi^*\circ s_0^*)+(\mathbb{1}- \pi^*\circ s_1^*)\\
& = -(-1)^{q-1}(dK_0-K_0d)+(-1)^{q-1}(dK_1-K_1d)\\
& = (-1)^{q-1}\big(dK_1-dK_0-K_1d+K_0d\big)\\
& = (-1)^{q-1}\big(d(K_1-K_0)-(K_1-K_0)d\big).
\end{align}
Now, recalling that if the homotopy map  is $F:M\times\mathbb{R}\to M$ ($F(x,0)=f(x)$, $F(x,1)=g(x)$), that
$$f^* = (F\circ s_0)^* = s_0^*\circ F^*$$
$$g^* = (F\circ s_1)^* = s_1^*\circ F^*$$
and also, that for any section $s$, $\pi\circ s = id$, the identity map on $M$, so that
$$s^*\circ\pi^* = (\pi\circ s)^* = (id)^* = \mathbb{1}$$
We see that
\begin{align}
f^*-g^* &= s_0^*\circ F^* - s_1^*\circ F^*\\
& = (s_0^* - s_1^*)\circ F^*\\
& = s^*\circ\pi^*\circ(s_0^* - s_1^*)\circ F^*\\
& = s^*\circ(-1)^{q-1}\big(d(K_1-K_0)-(K_1-K_0)d\big)\circ F^*\\
& = (-1)^{q-1}\bigg(d\Big(s^*\circ(K_1-K_0)\circ F^*\Big)-\Big(s^*\circ(K_1-K_0)\circ F^*\Big)d\bigg)
\end{align}
since pullbacks commute with the differential.
All this leads to an explicit form for a linear $\lambda$, namely
$$\lambda = s^*\circ(K_1-K_0)\circ F^*$$
where $F$ is the homotopy between $f$ and $g$, the $K$'s are integration over the fibres, and $s$ is any arbitrary section.
