Congruent iff Same Remainder (CISR) Confusion
Since $$0\le r_2<m$$ then \begin{align}&-m<-r_2\le0\quad \text{and since}\\ &\;\;\;0\le r_1<m\end{align} then by adding term by term we find $$-m<r_1-r_2<m$$
Sami explained the inequality. I explain another way, avoiding it.
$\qquad m\mid a\!-\!b \iff a\!-\!b \in m\Bbb Z \iff a+m\Bbb Z\, =\, b+m\Bbb Z$
$\qquad a\ {\rm mod}\ m\, $ is the least nonnegative element of $\ a+m\Bbb Z$
$\qquad b\ {\rm mod}\ m\, $ is the least nonnegative element of $\ b+m\Bbb Z.$
Being equal sets, they have equal least nonnegative elements.