Taking modulo of both sides On the following page at the bottom there is an algorithm for calculating modular inverses. In the proof I am confused with the line 'Taking both sides modulo $m$'. How does that work getting a congruence from the above equation?
$j = k\,\Rightarrow\, \color{#0a0}{j\bmod m} = k\bmod m\,$ by $\,f(x) = x\bmod m\,$ is a function so  $\,j=k\,\Rightarrow\, f(j) = f(k)$.

Your special case has $\, \color{#c00}0 \le \underbrace{m\bmod i}_{\Large\color{#0a0} j} &lt; i &lt; \color{#c00}m\,$ so $\,\color{#0a0}{j\bmod m = j}\,$ is already reduced $\!\bmod \color{#c00}m$

Remark $ $ Note that the modular inversion algorithm discussed there is the same as the algorithm I discussed in your prior question, i.e. Gauss's modular inversion algorithm (in non-fractional form).
If $x=y$ then $x \equiv y (mod m)$, so far, so good.

Now, let's say we have what I think you may be struggling with $ax + bm = y$

Then $ax + bm \equiv y\space (mod m)$, 

but $bm \equiv 0 \space (mod m)$, so that's why we get $ax \equiv y \space (mod m)$ To get there, you may want to subtract $bm$ from both sides