Understanding $\pmatrix{1 & 2 \\ 3 & 4} \pmatrix{2 & 0 \\ 1 & 2} ≠ \pmatrix{2 & 0 \\ 1 & 2} \pmatrix{1 & 2 \\ 3 & 4}$  Source
https://www.mathsisfun.com/algebra/matrix-multiplying.html
Seeking help if anyone could explain what is going under the hood.
How multiplication of x component with y component relates. It is said that you cannot compare apples with oranges but this multiplication of x component with y component appears to contradict. So while it is understood 1x2 (1x2+2x1) of the first row, first column as multiplication between x components, how 2 (x component) x 1 (y component) relates. It appears dot multiplication but an explanation will help.
 A: For me, it's good to follow the way mathematicians have hopefully discovered it (Cayley in particular in the 1850s) by considering the matrix product as an operation :
$$\binom{x}{y} \xrightarrow{A=\begin{pmatrix}a&b\\c&d\end{pmatrix}} \binom{p}{q} \xrightarrow{B=\begin{pmatrix}e&f\\g&h\end{pmatrix}} \binom{u}{v}\tag{0}$$
We are going to suppress the intermediate variables $p,q$ by expressing directly $u,v$ as (linear) expressions in variables $x,y$.
(0) means that
$$\binom{p}{q}=\begin{pmatrix}a&b\\c&d\end{pmatrix}\binom{x}{y} \iff \begin{cases}p&=&ax+by\\q&=&cx+dy\end{cases}\tag{1}$$
and
$$\binom{u}{v}=\begin{pmatrix}e&f\\g&h\end{pmatrix}\binom{p}{q} \iff \begin{cases}u&=&ep+fq\\v&=&gp+hq\end{cases}\tag{2}$$
Now it remains to plug relationships (1) into relationships (2) to get:
$$\begin{cases}u&=&e(ax+by)+f(cx+dy)\\v&=&g(ax+by)+h(cx+dy)\end{cases} \iff $$
$$\begin{cases}u&=&\color{red}{(ea+fc)}x+\color{red}{(eb+fd)}y\\v&=&\color{red}{(ga+hc)}x+\color{red}{(gb+hd)}y\end{cases},$$
where we recognize in red the entries of matrix product $BA$ (not $AB$ because $A$ is applied first, then $B$: we proceed from right to left).
