Theorem 8.1 of this book says the isomorphism \begin{equation} H_D(C^*(\mathfrak U,\Omega^*))\simeq H_\text{dR}^*(M). \end{equation} In this proof we consider a map $r:\Omega^*(M)\rightarrow\Omega^*(U)\oplus\Omega^*(V)\subset C^*(\mathfrak U,\Omega^*)$. This induces a map in cohomology: $r^*: H_\text{dR}^*(M)\rightarrow H_D^*(C^*(\mathfrak U,\Omega^*))$.
A $q$-cochain $\alpha\in C^*(\mathfrak U,\Omega^*)$ has two components, $\alpha = \alpha_0 + \alpha_1$, where $\alpha_0\in K^{0,1}$ and $\alpha_1\in K^{1,q-1}$. The exactness of the Mayer-Vietoris sequence gives $\exists\beta$ s.t. $\delta\beta = \alpha_1$. Then $\alpha - D\beta$ has only the $(0,q)$-component.
We need to prove that $r^*$ is an isomorphism. To prove $r^*$ is injective, suppose $r(\omega) = D\phi$ for some cochain $\phi\in C^*(\mathfrak U,\Omega^*)$. By writing as $\phi = \phi' + D\phi''$, where $\phi'$ has only the top component, \begin{equation} r(\omega) = D\phi' = d\phi'. \end{equation}
I thought the "top component" means $\alpha$ such that $\alpha\in K^{0,*}$ and $\delta\alpha = 0$. Then the above equality is obtained. Tell me if my thought is incorrect..
P.S. Later in Proposition 8.8 $r(\omega) = D\phi$ with $\phi$ has only the top component. The book says $\delta\phi = 0$. How to obtain this conclusion?