You learned this a long time ago:
"Positive times positive is positive",
"Positive times negative is negative",
"Negative times positive is negative",
"Negative times negative is positive".
The quotient group is of order 2. $N$ is automatically normal because $G$ is abelian. By definition:
$aN = bN$ if $a^{-1}b \in N$, which happens if and only if either both $a,b \in N$ or $a,b \not\in N$.
The two cosets are, explicitly: $N$ and $-N$, and these can be regarded as $[1]$ and $[-1]$.
This uses properties of the real numbers, which you are expected to know. The relevant property, here, is:
$\forall x \in \Bbb R-\{0\},\ x^2 > 0$ (that is: $x^2 \in N$).
This shows that for ANY coset $xN$, we have $(xN)(xN) = x^2N = N$.
So our cosets can only have order 1 ($N$ itself), or order 2.
Now if $y \not\in N$ (that is $y < 0$, see below) we have:
$(-1)^{-1}y = (\frac{1}{-1})y = (-1)y = -y \in N$, so that $yN = (-1)N$ which shows that the only OTHER coset is: $(-1)N = -N\ $ (this uses the trichotomy rule of the order on the real numbers).
Best Answer
The elements of the quotient group $S_4/V$ are the cosets of $V$ in $S_4$. So first I'd determine what elements of $S_4$ are in each of those six cosets and give them labels. For example $(1,2)$ is in the coset $B=\{(1,2),(3,4),(1,3,2,4),(1,4,2,3)\}.$ Remember that the cosets are disjoint.
From there start the multiplication table. To multiply the cosets pick a representative from each and multlitply in $S_4$. The choice of element in the coset doesn't matter as $V$ is normal so the quotient is well defined. For example to find $B*B$ pick a representative element from each, in this case, say, $(1,2)*(1,2)=e.$ Since $e$ is in $V$ we find that $B*B=V$.