Proposition: The order of $Hg\in G/H$ divides the order of $g\in G$ when $H$ is normal in $G$ and $G$ is finite.
Proof Attempt(by contradiction):
Let $H \lhd G$ and $gH\in G/H$ where the order of $gH$ is $k$ in $G$. Thus, $(gH)^k=g^kH=H \Leftrightarrow g^k\in H$ and $\forall j \lt k, (gH)^j=g^jH \neq H$ which implies $g^j \notin H$. Suppose $\vert g\vert=n$ such that $g\in G$ so that $g^n=e \in H$ and $\forall m \lt n, g^n\neq e$. By the division algorithm $\exists q,r\in \Bbb{Z}$ such that $n=kq+r$ where $0\leq r \lt q,k,n$. Hence, if $r\gt 0$, $g^r\neq e$ and $g^n=g^{kq+r}=g^{kq}g^r=e\Rightarrow g^r=g^{-kq}$. Since $g^k \in H$, then $g^{-kq}=g^r \in H$. But, $r<k$ so $g^r \notin H$ unless $r=0$. So $r$ must be $0$ and that means $n=kq$ and $k$ divides $n$, as desired.
If $\phi:G_1\to G_2$ is a group homomorphism and $g\in G_1$ has order $n$ then $\phi(g)^n=\phi(g^n)=\phi(e_1)=e_2$ implying that the order of $\phi(g)$ divides $n$.
You can apply that here for group homomorpism $G\to G/H$ prescribed by $g\mapsto gH$.