Let $A$, $B$, $C$, $D$, $E$ be rectangular real matrices of appropriate sizes, and define a psd matrix $S$ by $$ S = CB AA'B'C' + DD' + EE'. $$
Question 1. How to write $S$ as a product $$ S := F_1F_2\ldots F_k $$ of block matrices such that each block of each $F_p$ is either a zero matrix, and identity matrix, or one of the input matrices $A,B,C,D,E$ (or their transposes) ? Ideally, the number of factors $k$ should be small and each $F_i$ will not be very large. Presumably, $k$ is at least $6$ (which is the max degree of $S$ seen as a polynomial in the input matrices).
For example, if $A$ and $B$ identity matrices and $E=0$, then it's easy to see that $$ S = CC' + DD' = F_1 F_2, \text{where }F_1=\begin{bmatrix}C & D\end{bmatrix},\text{ and }F_2=F_1'. $$
More generally, let let $Y_{1,1},\ldots,Y_{1,m_1},\ldots, Y_{n,1},\ldots,Y_{n,m_n}$ be rectangular matrices of appropriate sizes and define a psd matrix $S$ by $$ S := \sum_{i=1}^n S_i, \text{ with }S_i = X_i X_i', \, X_i = \prod_{j=1}^{m_i} Y_{i,j}, $$
Question 2. Is there a generic way to express $S$ as a product of linear factors as in Question 1 (where now $A,B,C,D,E$ are replaced by the $Y_{i,j}$'s in the requirements) ?
Presumably, the the number of factors required is at least $2\max_i m_i$ (which is the degree of $S$).
Edit: Solution to Question 1
Consider the block matrices $F_1,\ldots,F_6$ given by
$$ F_1 = \begin{bmatrix}C & D & E\end{bmatrix},\, F_2 = \begin{bmatrix}B & 0 & 0\\ 0 & I & 0 \\ 0& 0 & I\end{bmatrix},\, F_3 = \begin{bmatrix}A & 0 & 0\\ 0 & I & 0 \\ 0& 0 & I\end{bmatrix},\, F_4=F_1',\, F_5=F_2',\, F_6=F_3'. $$
Observe that, $$ F_1 F_2F_3 = \begin{bmatrix}C & D & E\end{bmatrix}\begin{bmatrix}BA & 0 & 0\\ 0 & I & 0 \\ 0& 0 & I\end{bmatrix} = \begin{bmatrix}CBA & D & E\end{bmatrix}. $$
We deduce that $S=F_1 F_2 F_3 F_4 F_5 F_6$. General problem can be solved similarly.