Why $a\equiv b\pmod n$ implies $a$ and $b$ have equal remainder when divided by $n$? How does $a\equiv b\pmod n$ implies $a$ and $b$ have the same remainder when divided by $n$?

I don't understand the huge jump from modular equivalence to having equal remainders.

I see by definition 
$a\equiv b\pmod n\,$ implies $n\mid (a-b)\,$ so $\,a-b=nk\,$ so $\,a=b+nk$.

But I do not see how this implies that $a$ and $b$ have the same remainder when divided by $n$.

 A: $\,\bar a\, =\, a\bmod n\, =\, a-pn\ $ and 
$\,\bar b\, =\, b \bmod n\, =\, b-qn\ $ for some integers $\,p,q$.  
Therefore if $\ n\mid a-b\ $ then $\,n\,$ divides $\,\bar a-\bar b\, =\, a-b + (q-p)n$
By definition $\ 0 \le \bar a,\bar b < n\,$
therefore $\ 0 \le |\bar a -\bar b| < n$
so we conclude  $\ \bar a - \bar b = 0,\,$ being divisible by $\,n.$
A: It may be helpful for you instead to think about the following.
Proposition:  Let $a, b, m$ be integers with $m \geq 2$.  Then,
$$a \equiv b \pmod{m} \qquad \text{if and only if} \qquad a = mk + b$$ for some $k \in \mathbb{Z}$.
With the proposition in tow, consider the remainder of $a - b$ when divided by $m$.
A: We prove $\ n\mid a-b \iff a\,$ and $\,b\,$ leave the same remainder when divided by $\,n$.
If $a$ and $b$ have the same remainder $r$ when divided by $n$, then $a = q_1n + r$ and $b = q_2n + r$. Subtracting yields $\,a - b = q_1n + r - (q_2n + r) = (q_1 - q_2)n$, which means $n \mid  (a-b)$.
Conversely suppose $n\mid  (a-b).$ Then $\,a = b+nk\,$ for some integer $\,k.\,$ By the division algorithm, $b = qn + r\,$ for some integers $q$ and $0 \leq r < n$. Putting this value of $b$ into above $a = b + nk = (qn + r) + nk = (q + k)n + r$, so $r$ is also the remainder of $a$ divided by $n$.
