The recipe for multiplication of (scalar) matrices
$$
(AB)_{i,j}=\sum_k A_{i,k}B_{k,j}\tag1
$$
is saying: to obtain the $(i,j)$ element of $AB$, form the dot product as you walk along row $i$ of $A$ while simultaneously walking down column $j$ of $B$. In other words,

**The element at row $i$, column $j$ of $AB$ is the product of row $i$ of $A$ with column $j$ of $B$.**

Using the notation $A_{i,\ast}$ to denote row $i$ of $A$ and $B_{\ast,j}$ to denote column $j$ of $B$, this can be restated symbolically as
$$(AB)_{i,j}=A_{i,\ast}B_{\ast, j}\ .\tag2$$

This all works provided you can match up any row of $A$ with any column of $B$, which means the number of columns of $A$ must equal the number of rows of $B$. (This is what it means for $A$ and $B$ to be conformable.)

You can generalize (1) to block matrices, i.e., the formula still works when every $A_{i,k}$ and $B_{k,j}$ is itself a matrix [and the subscripts enumerate rows and columns of blocks], provided every product on the RHS of (1) makes sense. This means the number of columns of $A_{i,k}$ must equal the number of rows of $B_{k,j}$.

**Example:** Partition matrix $A$ into $A:=\pmatrix{A_1 &A_2}$ and $B$ into $B:=\pmatrix{B_1\\ B_2}$, where #columns($A_1$) = #rows($B_1$), and #columns($A_2$) = #rows($B_2$). Then
$$
AB=\pmatrix{A_1 & A_2}\pmatrix{B_1\\ B_2}=A_1B_1 + A_2 B_2\ .\tag3
$$
To see this, consider the element at row $i$, column $j$ of $AB$. This element is computed as the product of row $i$ of $A$ with column $j$ of $B$. But row $i$ of $A$ is just row $i$ of $A_1$ followed by row $i$ of $A_2$. Similarly, column $j$ of $B$ is column $j$ of $B_1$ atop column $j$ of $B_2$. Assuming conformability, the product of row $i$ of $A$ with column $j$ of $B$ is the sum of two pieces: one piece is the product of row $i$ of $A_1$ with column $j$ of $B_1$, and the other is the product of row $i$ of $A_2$ with column $j$ of $B_2$. In symbols we've argued that
$$(AB)_{i,j}=A_{i,\ast}B_{\ast,j}=(A_1)_{i,\ast}(B_1)_{\ast,j}+(A_2)_{i,\ast}(B_2)_{\ast,j}=(A_1B_1)_{i,j} + (A_2B_2)_{i,j},\tag4$$ and we're done.

**Example:** Consider conformable partitioned matrices $C:=\pmatrix{C_1\\ C_2}$ and $D:=\pmatrix{D_1 & D_2}$. Then
$$CD=\pmatrix{C_1\\ C_2}\pmatrix{D_1 & D_2}
=\pmatrix{C_1D_1 & C_1D_2\\C_2D_1&C_2 D_2}.\tag5
$$ Again, this can be seen by considering what happens when you multiply row $i$ of $C$ with column $j$ of $D$.