Why I think you formula contains an error $\newcommand{\d}{\mathop{}\!d}$
Let $n = 2$ and $\omega = \d x \wedge \d y$. Then according to your formula
$$I \omega = x \left(\int_0^1 s \d s \right) \d y + y \left( \int_0^1 s \d s\right) \d x = \frac12 (x \d y + y \d x) $$
so $\d (I \omega) = \frac12(\d x \wedge \d y + \d y \wedge \d x) = 0$ and since $\d\omega = 0$ the equation you want to prove does not hold. However if you add a factor of $(-1)^{j-1}$ to the inner sum in the definition of $I \omega$ then $I \omega$ would be $\frac12(x \d y - y \d x)$ in this example and everything works out. The formula for $I$ then is
$$
(I\omega)
= \sum_{i_1\lt\dots\lt i_k}
\sum_{j=1}^kx_{i_j}
\left( \int_0^1s^{k-1}c_{i_1,\dots,i_k}(sx) \d s\right)
(-1)^{j-1} \d x_{i_1}\wedge\dots \wedge \d x_{i_{j-1}}\wedge \d x_{i_{j+1}}\wedge\dots\wedge \d x_{i_k}
$$
Proving $\d(I \omega) + I(\d \omega) = \omega$
Since $\d$ and $I$ are bot additive, it is sufficient to verify this for $\omega = f(x) \d x_{i_1} \wedge \dots \wedge \d x_{i_k}$. By further reordering of the coordinates we can assume $\omega = f \d x_1 \wedge \dots \wedge d x_k$.
Let's denote $\d x_1 \wedge \dots \wedge d x_k$ by $\d x_I$ and $\d x_1 \wedge \dots \wedge \d x_{j-1} \wedge \d x_{j+1} \wedge \dots \wedge d x_k$ by $\d x_{I - j}$.
We have
$$ I \omega = \sum_{j=1}^k x_j \left( \int_0^1 s^{k-1} f(s x) \d s \right) (-1)^{j-1} \d x_{I - j}
$$
and
$$
\begin{align}
\d (I \omega) &= \sum_{i = 1}^n \sum_{j=1}^k \frac{\partial}{\partial x_i} \left[x_j \int_0^1 s^{k-1} f(s x) \d s \right] \d x_i \wedge (-1)^{j-1} d x_{I - j} \\
&= \sum_{i=1}^k \frac{\partial}{\partial x_i} \left[x_i \int_0^1 s^{k-1} f(s x) \d s \right] \d x_I \\
&\phantom{=}+ \sum_{i = k + 1}^n \sum_{j=1}^k x_j \left( \int_0^1 s^k \frac{\partial f}{\partial x_i}(s x) \d s \right) (-1)^{j-1} (-1)^{k-1} d x_{I - j} \wedge d x_i \tag{1}
\end{align}
$$
We can simplify the first sum:
$$
\begin{align}
\sum_{i=1}^k \frac{\partial}{\partial x_i} \left[x_i \int_0^1 s^{k-1} f(s x) \d s \right] &=
\int_0^1 k s^{k-1} f(s x) \d s + \int_0^1 s^k \sum_{i=1}^k x_i \frac{\partial f}{\partial x_i} (s x)
\end{align}
$$
Now
$$
\d \omega = \sum_{i = k + 1}^n \frac{\partial f}{\partial x_i} \d x_i \wedge \d x_I
= (-1)^k \sum_{i = k + 1}^n \frac{\partial f}{\partial x_i} \d x_I \wedge \d x_i
$$
and
$$
\begin{align*}
I(\d \omega) = (-1)^k \sum_{i = k+1}^n &\left[ \sum_{j=1}^k x_j \left( \int_0^1 s^k \frac{\partial f}{\partial x_i}(s x) \right) (-1)^{j-1} \d x_{I - j} \wedge \d x_i \right.\\
&\phantom{[}\left. + x_i \left( \int_0^1 s^k \frac{\partial f}{\partial x_i}(s x) \right) (-1)^k \d x_I\right] \tag{2}
\end{align*}
$$
The first sum in $(2)$ is exactly the negative of the second sum in $(1)$ - $(-1)^k$ vs $(-1)^{k-1}$ - so they cancel when adding $I(\d \omega)$ and $\d (I \omega)$ What remains is
$$
\left( \int_0^1 k s^{k-1} f(s x) \d s + \int_0^1 s^k \sum_{i=1}^n x_i \frac{\partial f}{\partial x_i} (s x) \right) \d x_I
$$
Since the sum on the right hand side is the total derivative of $f(s x)$ with respect to $s$, partial integration shows that this is equal to $f(x) \d x_I = \omega$.
Let $S$ be that surface and $\iota : S\to U$ be the inclusion. The fact that $\omega$ vanishes over the tangent vectors of $S$ is the same as saying that $\iota^*\omega = 0$.
With that in mind, the proof is trivial: Since pullback $\iota^*$ commutes with exterior differentiation $d$,
$$ \iota^* (d\omega) = d (\iota^* \omega) = d(0) = 0.$$
Best Answer
For concreteness, let's take $n = 4$, so the coordinates on $\mathbb{R}^4$ are $(x_1, x_2, x_3, x_4)$.
$\bullet$ A differential $1$-form $\omega$ on an open set $U \subset \mathbb{R}^4$ can be written $$\omega = \sum_{1 \leq i \leq 4} f_i\,dx_i = f_1dx_1 + f_2\,dx_2 + f_3\,dx_3 + f_4dx_4$$ for some functions $f_1, \ldots, f_4 \colon U \to \mathbb{R}$.
$\bullet$ A differential $2$-form $\omega$ on an open set $U \subset \mathbb{R}^4$ can be written \begin{align*} \omega & = \sum_{1 \leq i < j \leq 4} f_{ij}\,dx_i \wedge dx_j \\ & = f_{12}\,dx_1 \wedge dx_2 + f_{13}\,dx_1 \wedge dx_3 + f_{14}\,dx_1 \wedge dx_4 \\ & \ \ \ \ \ \ \ + f_{23}\,dx_2 \wedge dx_3 + f_{24} \,dx_2 \wedge dx_4 + f_{34}\,dx_3 \wedge dx_4 \end{align*} for some functions $f_{12}, f_{13}, f_{14}, f_{23}, f_{24}, f_{34} \colon U \to \mathbb{R}$.
$\bullet$ A differential $3$-form $\omega$ on an open set $U \subset \mathbb{R}^4$ can be written \begin{align*} \omega & = \sum_{1 \leq i < j < m \leq 4} f_{ijm}\,dx_i \wedge dx_j \wedge dx_{m} \\ & = f_{123}\,dx_1 \wedge dx_2 \wedge dx_3 + f_{124}\,dx_1 \wedge dx_2 \wedge dx_4 \\ & \ \ \ \ \ + f_{134}\,dx_1 \wedge dx_3 \wedge dx_4 + f_{234}\,dx_2 \wedge dx_3 \wedge dx_4 \end{align*} for some functions $f_{123}, f_{124}, f_{134}, f_{234} \colon U \to \mathbb{R}$.
$\bullet$ A differential $4$-form $\omega$ on an open set $U \subset \mathbb{R}^4$ can be written \begin{align*} \omega & = \sum_{1 \leq i < j < m < p \leq 4} f_{ijmp}\,dx_i \wedge dx_j \wedge dx_{m} \wedge dx_p = f_{1234} \,dx_1 \wedge dx_2 \wedge dx_3 \wedge dx_4 \end{align*} for some function $f_{1234} \colon U \to \mathbb{R}$.
In general, when we work with differential $k$-forms on $\mathbb{R}^n$, we have to resort to writing the summation range as $$1 \leq i_1 < i_2 < \cdots < i_k \leq n$$ for the simple reason that we don't have infinitely many letters in the alphabet.