It's the so called Poincare's theorem from Rudin's book. I read this theorem fully but I have 2 questions:
1) How did he conclude that equations (120) holds? What did he use in his reasoning? This seems to me a bit weird.
2) He wrote that $d^2\beta=0$. Why? I guess that's why $\beta \in C''$. To prove strictly that $\beta\in C''$ we have no information about $(D_kF_I)(\mathbf{x})$ for $1\leqslant k<p$.
Any answer would be appreciated.
Let's start with equation $(120)$. We have two assumptions, first that $\omega$ is closed, $d\omega = 0$, and second that $\omega \in Y_p$, that is, in
$$\omega = \sum_I f_I(x)\,dx_I$$ all $k$-indices $I$ only involve $dx_i$ for $i \leqslant p$. We compute
$$d\omega = \sum_I (d f_I)(x)\wedge dx_I = \sum_I \sum_{r = 1}^n (D_r f_I)(x)\, dx_r \wedge dx_I = \sum_{r = 1}^n \sum_I (D_r f_I)(x)\,dx_r \wedge dx_I.$$
Now we split the outer sum in two parts, one with $r \leqslant p$ and the other with $r > p$:
$$d\omega = \underbrace{\sum_{r = 1}^p \sum_I (D_r f_I)(x)\,dx_r\wedge dx_I}_{A} + \underbrace{\sum_{r = p+1}^n \sum_I (D_r f_I)(x)\, dx_r \wedge dx_I}_{B}$$
In part $A$, only differentials $dx_i$ with $i \leqslant p$ occur, and each term in $B$ contains exactly one differential $dx_i$ with $i > p$. Thus there is no cancelling between terms in $A$ and terms in $B$, Since $(I_1 \cup \{r_1\}) = (I_2 \cup \{r_2\})$ for $k$-indices $I_1, I_2 \subset \{1,\dotsc, p\}$ and $r_1, r_2 \in \{p+1, \dotsc,n\}$ only holds when $I_1 = I_2$ and $r_1 = r_2$, we have no cancelling between any terms in $B$ either, and thus $d\omega = 0$ and $\omega\in Y_p$ together imply
$$D_r f_I \equiv 0 \tag{120}$$
for $p < r \leqslant n$.
This means that none of the coefficients of $\omega$ depends on any of the $x_i$ with $i > p$. We can thus, for convenience of notation assume $n = p$.
Now, coming to $\beta$, I will use a slightly different notation from Rudin's, I write
$$\omega = \underbrace{\sum_I f_I(x',x_p)\,dx_I}_{\alpha} + \sum_K g_K(x',x_p) \,dx_p \wedge dx_K,\tag{121'}$$
where $I$ ranges over the increasing $k$-indices in $\{1,\dotsc, p-1\}$ and $K$ ranges over the increasing $(k-1)$-indices in $\{1,\dotsc,p-1\}$, and $x' = (x_1,\dotsc,x_{p-1})$. My $K$ is Rudin's $I_0$, and my $g_K$ is $(-1)^{k-1}$ times Rudin's $f_I$ where $I = (K,p)$. The factor $(-1)^{k-1}$ comes from $dx_p \wedge dx_K = (-1)^{k-1} dx_k \wedge dx_p$ of course.
The closedness of $\omega$ implies some relations between derivatives of the $f_I$ and the $g_K$, namely
$$D_p f_I + \sum_{m = 1}^k (-1)^m D_{i_m} g_{I - i_m} = 0, \tag{$\ast$}$$
where $I - i_m$ is the increasing $(k-1)$-index obtained from $I$ be removing the index $i_m$. The left hand side of $(\ast)$ is the coefficient of $dx_p \wedge dx_I$ in $d\omega = 0$.
Next, we use theorem 10.38 to define $\mathscr{C}'$ functions $G_K$ with $D_p G_K = g_K$, explicitly
$$G_K(x',x_p) = \int_{\varphi(x')}^{x_p} g_K(x',t)\,dt,$$
where $\varphi$ is a $\mathscr{C}''$ function defined on $E' = \bigl\{ x' : (\exists t)\bigl((x',t) \in E\bigr)\bigr\}$ with $(x',\varphi(x')) \in E$ for all $x' \in E'$. Rudin only requires the function - which is called $\alpha$ in 10.38, but here $\alpha$ denotes something else - to be of class $\mathscr{C}'$, which suffices to have $\beta$ a $\mathscr{C}'$ form, but in order to guarantee that $d\beta$ is also of class $\mathscr{C}'$, I need $\varphi$ twice continuously differentiable. One can always have a smooth (infinitely often differentiable) such function.
Now we define
$$\beta = \sum_K G_K(x',x_p) \,dx_K.$$
This differs from Rudin's $\beta$ by a factor of $(-1)^{k-1}$.
To compute $d\beta$, we need the partial derivatives of $G_K$ with respect to the $x_r$ for $r < p$. By Leibniz' rule, we have
$$D_r G_K (x',x_p) = \int_{\varphi(x')}^{x_p} D_r g_K(x',t)\,dt - g_K(x',\varphi(x'))\cdot D_r\varphi(x')$$
for $r < p$. Here we see why we need $\varphi$ to be twice continuously differentiable, when computing $d(d\beta)$ we must also differentiate $D_r\varphi$. As a first step we get
\begin{align} d\beta &= \sum_K D_p G_K(x',x_p) \,dx_p\wedge dx_K + \sum_K \sum_{r = 1}^{p-1} D_r G_K(x',x_p) \,dx_r \wedge dx_K\\ &= \sum_K g_K(x',x_p)\,dx_p \wedge dx_K + \sum_K \sum_{r = 1}^{p-1} \Biggl(\int_{\varphi(x')}^{x_p} D_r g_K(x',t)\,dt - g_K(x',\varphi(x'))D_r\varphi(x')\Biggr) dx_r \wedge dx_K. \end{align}
In the second sum, only differentials $dx_i$ with $i < p$ occur, and when $r$ occurs in $K$, then $dx_r \wedge dx_K = 0$, while when $r$ doesn't occur in $K$, then $K \cup \{r\}$ corresponds to a unique increasing $k$-index in $\{1,\dotsc,p-1\}$, and each of these is reached from $k$ pairs $(K,r)$. We can thus write
\begin{align} \sum_K \sum_{r = 1}^{p-1} \Biggl(\int_{\varphi(x')}^{x_p} D_r g_K(x',t)\,dt\Biggr) \,dx_r \wedge dx_K &= \sum_I \sum_{m = 1}^k (-1)^{m-1} \Biggl(\int_{\varphi(x')}^{x_p} D_{i_m} g_{I-i_m} (x',t)\,dt\Biggr)\,dx_I\\ &= \sum_I \Biggl(\int_{\varphi(x')}^{x_p} \sum_{m = 1}^k (-1)^{m-1} D_{i_m} g_{I-i_m}(x',t)\,dt\Biggr) dx_I\\ &= \sum_I \Biggl( \int_{\varphi(x')}^{x_p} D_p f_I(x',t)\,dt\Biggr)dx_I \tag{by $(\ast)$}\\ &= \sum_I f_I(x',x_p)\,dx_I - \sum_I f_I(x',\varphi(x'))\,dx_I. \end{align}
Thus we arrive at
$$d\beta = \omega - \sum_I f_I(x',\varphi(x'))\,dx_I - \sum_K \sum_{r = 1}^{p-1} g_K(x',\varphi(x'))D_r\varphi(x')\,dx_r \wedge dx_K.$$
Here it is easy to see that $d\beta\in \mathscr{C}'$, and since $d\omega = 0$, $d^2\beta = 0$ is equivalent to $d\gamma = 0$, where
$$\gamma = \omega - d\beta = \underbrace{\sum_I f_I(x',\varphi(x'))\,dx_I}_{\gamma_1} + \underbrace{\sum_K \sum_{r = 1}^{p-1} g_K(x',\varphi(x'))D_r\varphi(x')\,dx_r\wedge dx_K}_{\gamma_2}.$$
Now to see that $d\gamma = 0$, we group the terms of $d\gamma$ by the derivatives of $\varphi$ that occur. We get a second derivative of $\varphi$ only from $\gamma_2$. This yields the contribution
$$\sum_K g_K(x',\varphi(x'))\Biggl(\sum_{r = 1}^{p-1} \sum_{s = 1}^{p-1} D_sD_r\varphi(x') \,dx_s\wedge dx_r\Biggr) \wedge dx_K.$$
Since $\varphi$ is twice continuously differentiable, we have $D_sD_r\varphi = D_rD_s\varphi$, and since $dx_s \wedge dx_r = - dx_r \wedge dx_s$, the inner double sum cancels completely for every $K$, so that contribution is $0$. Next, we get to first derivatives of $\varphi$ also only from $\gamma_2$, as the second term from
$$\frac{\partial}{\partial x_s} g_K(x',\varphi(x')) = D_s g_K(x',\varphi(x')) + D_pg_K(x',\varphi(x'))D_s\varphi(x').$$
This yields the contribution
$$\sum_K D_pg_K(x',\varphi(x'))\Biggl(\sum_{s = 1}^{p-1}\sum_{r = 1}^{p-1} D_s\varphi(x')D_r\varphi(x') \,dx_s \wedge dx_r\Biggr)dx_K$$
which again vanishes since $dx_s \wedge dx_r + dx_r \wedge dx_s = 0$.
The third part will consist of the terms in which no derivative of $\varphi$ occurs at all. These come from $\gamma_1$, and the contribution is
$$\sum_I \sum_{r = 1}^{p-1} D_r f_I(x',\varphi(x'))\,dx_r \wedge dx_I.$$
This is just the part of $d\omega$ in which $dx_p$ doesn't occur, evaluated at $(x',\varphi(x'))$, and since $d\omega = 0$, it follows that this also vanishes.
What remains are the terms in which one first derivative of $\varphi$ occurs. From $\gamma_1$, we get
$$\sum_{r = 1}^{p-1}\sum_I D_p f_I(x',\varphi(x')) D_r\varphi(x')\,dx_r\wedge dx_I$$
and from $\gamma_2$ we get
$$\sum_{r = 1}^{p-1} \sum_K \sum_{s = 1}^{p-1} D_s g_K(x',\varphi(x')) D_r\varphi(x')\,dx_s \wedge dx_r \wedge dx_K.$$
Fixing $r$, we need
$$\sum_I D_p f_I(x',\varphi(x'))\,dx_I - \sum_K \sum_{s = 1}^{p-1} D_s g_K(x',\varphi(x'))\, dx_s \wedge dx_K = 0.$$
This again follows from $(\ast)$, since $dx_s \wedge dx_K = 0$ when $s$ occurs in $K$, and sorting $s$ into $K$ to obtain an increasing $k$-index when $s$ doesn't occur in $K$.
Finally, since $\gamma \in Y_{p-1}$, the induction hypothesis yields $\gamma = d\mu$ and therefore $\omega = d\beta + \gamma = d\beta + d\mu = d(\beta + \mu)$.