This is in John Lee's Smooth Manifold 2nd Edition, pg 444
For any smooth manifold $M$, there exists a linear map $$ h:\Omega^p(M \times I ) \rightarrow \Omega^{p-1}(M)$$ such that $$ h(dw)+d(hw) = i_1^*w - i^*_0w$$
The proof begins goes as follows. This is the same one here.
Let $s$ be the standard coordinate on $\Bbb R$ and let $S$ be the vector field on $M \times \Bbb R$ given by $S_{(q,s)} = (0, \partial/\partial s|_s)$. Let $w$ be a smooth $p$-form on $M \times I$. Define $hw \in \Omega^{p-1}(M)$ by $$ hw = \int_0^1 i^*_t (S \lrcorner w) \, dt. $$ Specifically, given $q \in M$, this means, $$ (hw)_q = \int_0^1 i^*_t\Big( (S \lrcorner w )_{(q,t)} \Big) \, dt $$
The notation $S \lrcorner w$ is interior multiplication. Otherwise denoted by $\iota_Sw$.
My question is, what does this integral even mean?
At each point $q \in M$, the integrand is a function of $t$ with values in the vector space $\wedge^{p-1} (T^*q M)$. How do we integrate over this?
My thoughts on the way to see this:
We work locally, in nhood $U_q$. For $r \in U_q$ our integrand is given by $$ \sum a_I(r,t) dx^I$$ summing over indices $I$ increasing $p-1$ length subset of $0$ to $n$. Our new $p-1$ form is given by $$ \sum \int a_I(r,t) \, dt \, d x^I $$
There is a problem however, that we have to show this is independent of the choice of coordinate.
