How to multiply a matrix of block matrices?

Your multiplication is fine; the problem is a typo in the notes. They should read $$\left[\begin{matrix} A_{m\times m}&B_{m\times n}\\ C_{n\times n}&D_{n\times n} \end{matrix}\right]= \left[\begin{matrix} A_{m\times m}&0_{m\times n}\\ C_{n\times m}&I_{n\times n} \end{matrix}\right] \left[\begin{matrix} I_{m\times m}&A^{-1}B_{m\times n}\\ 0_{n\times m}&D-CA^{-1}B_{n\times n} \end{matrix}\right], $$ with a $0_{n\times m}$ in the lower left-hand corner of the second factor, just as the second factor in the next line of the notes has $0_{m\times n}$ in the upper right-hand corner.

Perhaps someone else has a feel for where the decomposition comes from; I’d be hard pressed to come up with it and can merely observe after the fact that it works.

This is a typo in the proof. The second factor should be

$$\begin{bmatrix} I_{m \times m} & A^{-1}B_{m \times n}\\ 0_{n \times m} & D-CA^{-1}B_{n \times n} \end{bmatrix}$$

with a zero instead of a $C$, in analogy with the other product. Note that the claimed determinant wouldn't come out right either with the $C$ included.

The decomposition is related to the Schur complement; see also A geometric way to reason about Schur complements?