Let $a\in G$ have finite order; it can be computed as the minimum $k>0$ such that $a^k=1$.
But the same holds for an element $aN\in G/N$: the order is the minimum exponent $h>0$ such that $(aN)^h=1N$, which means the minimum $h>0$ such that $a^h\in N$, because $(aN)^h=a^hN$ and $gN=1N$ if and only if $g\in N$.
Thus $a^k=1$ certainly implies that $a^k\in N$, but the converse may not be true.
If you use a different characterization of the order, namely the cardinality of the cyclic subgroup generated by the given element, you can be even more precise.
Consider the canonical projection $\pi\colon G\to G/N$. Then, certainly,
$$
\pi(\langle a\rangle)=\langle aN\rangle
$$
so, by the homomorphism theorems,
$$
\langle aN\rangle \cong \frac{\langle a\rangle N}{N}\cong
\frac{\langle a\rangle}{\langle a\rangle \cap N}
$$
which implies that the order of $aN$ divides the order of $a$. The examples you find in other answers show that any divisor can result.
In the case when $a$ has infinite order, the above reasoning with the homomorphism theorem still holds, showing that the order of $aN$ can be any integer (or be infinite).
So what the order of $aN$ is strictly depends on both $a$ and $N$ and nothing more than “the order of $aN$ divides the order of $a$” can be said in general.