Prove that: If $p$ is an odd prime, then $(2/p)= \begin{cases} 1, & \text{if $p = 1 (\mod 8)$ or $7 (\mod 8)$} \\ -1, & \text{if $p = 3 (\mod 8)$ or $5 (\mod 8)$}\end{cases}$
According to Gauss' lemma, $(2,p) = (-1)^{n}$, where $n$ is the number of integers in the set $$S = \{1 * 2, 2 * 2, \dots, (\frac{p-1}{2}) * 2 \} $$ which, upon division by $p$, have remainders greater than $p/2$. The members of $S$ are all less than $p$, so that it suffices to count the number that exceeds $p/2$
Why does this suffice? I can't see how that if $2k \lt p$ and $2k \gt \frac{p}{2}$ then $\frac{2k}{p} \gt \frac{p}{2}$.
By "upon division by $p$, have remainders...", it is meant that $S$ is the set of positive residues $2i$ mod $p$.
But for your case, all the values $(2i)$ are less than $p$, so we don't need to do anything special to find these residues; we can just look at the values.
So "it suffices to count the values $2i$ that exceed $p/2$".
Note that if we were looking at $(3,p)$ instead, we couldn't act so simply; because (for example) $3\frac{p-1}{2}$ is greater than $p$, so we must evaluate it mod $p$; and only then determine whether that result is $> p/2$.