I've seen similar posts here asking why $(ab)\bmod m = (a \bmod m)( b \bmod m)$, but none of them really answers my question or gives a detailed the mathematical proof.
I know that $a\equiv b \pmod m$, $c\equiv d \pmod m)$ implies $ac\equiv bd \pmod m$, and I was going to approach this by setting $a = k_am + a \pmod m$ and $b = k_bm + b \pmod m$ then multiply $a$ and $b$, but I got stuck.
It is a definition, you want define a product on $\mathbb{Z}/m\mathbb{Z}$ but you have only a product on $\mathbb{Z}$. A way to define product on $\mathbb{Z}/m\mathbb{Z}$ it is to prove that
if $a\cong c$ mod $m $ and $b\cong d$ mod m then
$ab\cong cd $ mod $m$ so it is well define the following operation on $\mathbb{Z}/m\mathbb{Z}$ :
($a$ mod $m$)($b$ mod $m$):= $ab$ mod $m$$