As mentioned, the definition of normal subgroup specifies that
$g Ng^{-1} = N \tag 1$
for all $g \in G$. We note this entails
$\forall g \in G, \; g^{-1}N g = g^{-1}N(g^{-1})^{-1} = N; \tag 2$
therefore, though the mappings
$g \mapsto gng^{-1} \tag 3$
and
$g \mapsto g^{-1}ng \tag 4$
are not in general the same, they give rise to the same definition of normal subgroup, since conjugation by every $g \in G$ is required; thus we always pick up both left and right conjugation by any paricular $g$. So in fact the choice of right or left conjugation is arbitrary when defining normal subgroup.
In closing, note that left and right conjugation are the same, that is,
$gng^{-1} = g^{-1}ng \tag 5$
if and only if
$g^2n = ng^2, \; \forall n \in N, \; g \in G; \tag 6$
i.e., when ever every element of $N$ commutes with every square in $G$, when the set of squares is contained in $C_N(G)$, the centralizer of $N$ in $G$.
Here is @Arturo Magidin's comment that answers my question. I post it here to remove this question from unanswered list. All credits are given to @Arturo Magidin.
In the second one, you can’t “substitute” anything for $N$. You can pick the element you want, but you cannot replace $N$ with something else. Instead, once you have $g^{-1}Ng\subseteq N$, multiply on the left by $g$ and on the right by $g^{-1}$.
Best Answer
So that you don't need to prove the reverse inclusion each time you want to test whether a group is normal.
That is, if $N$ is a subgroup of $G$, and I show that for all $n\in N$, $g\in G$, we have that $gng^{-1}\in N$, then $N$ is normal to $G$ automatically.
If we only relied on the third condition, we would have to further prove that for all $n\in N$, $g\in G$, $n\in gNg^{-1}$. Due to the statement given, we see that we automatically get this from the second condition.