I'm trying to come up with a group $G$ and normal subgroup $H$ of $G$ such that $G/H$ contains an element of order $n$ (for some integer $n$), but $G$ does not.
Does $G = \mathbb{Z}$ and $H = 5\mathbb{Z}$ work? $G/H = \mathbb{Z}/5\mathbb{Z}$ contains an element of order $4$ but $\mathbb{Z}$ does not (both groups viewed under addition, of course).
Could someone confirm for me?
If you meant $\mathbb{Z}/5\mathbb{Z}$ contains an element of order five, then you are correct.
The element $1 + 5\mathbb{Z} \in \mathbb{Z}/5\mathbb{Z}$ has order five (as does any other non-zero element) but $\mathbb{Z}$ does not - every non-zero element of $\mathbb{Z}$ has infinite order. If you prefer to think of $\mathbb{Z}/5\mathbb{Z}$ as $\{0, 1, 2, 3, 4\}$ with group operation addition modulo $5$, then $1$ is an element of order five; in fact, under the natural isomorphism $\mathbb{Z}/5\mathbb{Z} \to \{0, 1, 2, 3, 4\}$, $1 + 5\mathbb{Z} \mapsto 1$.
Note, there are infinitely many other examples of this type. For any $n \in \mathbb{N}$, $\mathbb{Z}/n\mathbb{Z}$ has an element of order $n$, namely $1 + n\mathbb{Z}$, but $\mathbb{Z}$ does not.