In this proposition the homotopy operator $L: C^*(\mathfrak U,\Omega^*)\rightarrow C^*(\mathfrak U,\Omega^*)$, \begin{equation} L\alpha = \sum_{p=0}^{n-1}(L\alpha)_p,\;\; (L\alpha)_p = \sum_{i=p+1}^{n}K(-D''K)^{i-(p+1)}\alpha_i\;\in C^p(\mathfrak U,\Omega^{n-(p+1)}) \end{equation} is defined.
For $\alpha = \sum_{i=0}^{n}\alpha_i$ and $D\alpha = \beta = \sum_{i=0}^{n+1}\beta_i$, we define \begin{equation} f(\alpha) := \sum_{i=0}^{n}(-D''K)^i\alpha_i - \sum_{i=0}^{n}K(-D''K)^i\beta_{i+1}\;\in C^0(\mathfrak U,\Omega^n). \end{equation}
The exercise to confirm the relation $1-r\circ f = DL + LD$ is left to the readers. How can we prove it?
The following is my procedure. I think I have almost done but still have minor mistakes. Please check..
\begin{align*} DL(\alpha) + LD(\alpha) &= (\delta+D'')\sum_{p=0}^{n-1}\sum_{i=p+1}^{n}K(-D''K)^{i-(p+1)}\alpha_i + L\beta\\ &= \sum_{p=0}^{n-1}\sum_{i=p+1}^{n}(-D''K)^{i-(p+1)}\alpha_i - \sum_{p=0}^{n-1}\sum_{i=p+1}^{n}K\delta(-D''K)^{i-(p+1)}\alpha_i\\ &\qquad - \sum_{p=0}^{n-1}\sum_{i=p+1}^{n}(-D''K)^{i-p}\alpha_i + \sum_{p=0}^{n}\sum_{i=p+1}^{n+1}K(-D''K)^{i-(p+1)}\beta_i\\ &= \sum_{p=0}^{n-1}\sum_{i=p+1}^{n}(-D''K)^{i-(p+1)}\alpha_i - \sum_{p=0}^{n-1}\sum_{i=p+1}^{n}K(-D''K)^{i-(p+1)}\delta\alpha_i - \sum_{p=0}^{n-1}\sum_{i=p+1}^{n}K(-D''K)^{i-(p+2)}D''\alpha_i\\ &\qquad - \sum_{p=0}^{n-1}\sum_{i=p+1}^{n}(-D''K)^{i-p}\alpha_i + \sum_{p=0}^{n-1}\sum_{i=p+1}^{n+1}K(-D''K)^{i-(p+1)}\beta_i\\ &= \sum_{p=0}^{n-1}\sum_{i=p+1}^{n}(-D''K)^{i-(p+1)}\alpha_i - \sum_{p=0}^{n-1}\sum_{i=p+1}^{n}K(-D''K)^{i-(p+1)}(-D''\alpha_{i+1}+\beta_{i+1}) - \sum_{p=0}^{n-1}\sum_{i=p+1}^{n}K(-D''K)^{i-(p+2)}D''\alpha_i\\ &\qquad - \sum_{p=0}^{n-1}\sum_{i=p+1}^{n}(-D''K)^{i-p}\alpha_i + \sum_{p=0}^{n-1}\sum_{i=p+1}^{n+1}K(-D''K)^{i-(p+1)}\beta_i\\ &= \sum_{p=0}^{n-1}\sum_{i=p+1}^{n}(-D''K)^{i-(p+1)}\alpha_i - \sum_{p=0}^{n-1}\sum_{i=p+2}^{n+1}K(-D''K)^{i-(p+2)}(-D''\alpha_i+\beta_i) - \sum_{p=0}^{n-1}\sum_{i=p+1}^{n}K(-D''K)^{i-(p+2)}D''\alpha_i\\ &\qquad - \sum_{p=0}^{n-1}\sum_{i=p+1}^{n}(-D''K)^{i-p}\alpha_i + \sum_{p=0}^{n-1}\sum_{i=p+1}^{n+1}K(-D''K)^{i-(p+1)}\beta_i\\ &= \sum_{p=0}^{n-1}\sum_{i=p+1}^{n}(-D''K)^{i-(p+1)}\alpha_i - \sum_{p=0}^{n-1}\sum_{i=p+1}^{n}K(-D''K)^{i-(p+1)}\beta_{i+1}\\ &\qquad - \sum_{p=0}^{n-1}\sum_{i=p+1}^{n}(-D''K)^{i-p}\alpha_i + \sum_{p=0}^{n-1}\sum_{i=p}^{n}K(-D''K)^{i-p}\beta_{i+1}. \end{align*} Changing of the ordering of the sum gives for $\alpha_i$ parts, \begin{equation*} \sum_{i=0}^{n}\sum_{p=0}^{i-1}(-D''K)^{i-(p+1)}\alpha_i - \sum_{i=0}^{n}\sum_{p=0}^{i-1}(-D''K)^{i-p}\alpha_i = \sum_{i=0}^{n}\sum_{p=1}^{i}(-D''K)^{i-p}\alpha_i - \sum_{i=0}^{n}\sum_{p=0}^{i-1}(-D''K)^{i-p}\alpha_i = \sum_{i=0}^{n}\alpha_i - \sum_{i=0}^{n}(-D''K)^i\alpha_i. \end{equation*} For $\beta_i$ parts, \begin{align*} -\sum_{i=1}^{n}\sum_{p=0}^{i-1}K(-D''K)^{i-(p+1)}\beta_{i+1} + \sum_{i=0}^{n}\sum_{p=0}^{i}K(-D''K)^{i-p}\beta_{i+1} &= -\sum_{i=1}^{n}\sum_{p=1}^{i}K(-D''K)^{i-p}\beta_{i+1} + \sum_{i=0}^{n}\sum_{p=0}^{i}K(-D''K)^{i-p}\beta_{i+1}\\ &= \sum_{i=0}^{n}K(-D''K)^i\beta_{i+1}. \end{align*} By adding them, \begin{equation} (DL+LD)(\alpha) = \alpha - r\circ f(\alpha). \end{equation}