$\textbf{Proposition:}$ Let $G = \langle x_0\rangle$ be a cyclic group of order $n\in \mathbb{N}_0$ and $H$ a finite group. Then the number of group homomorphisms $\varphi:G\to H$ is equal to the number of elements $h\in H$ such that $\text{order}(h)|n$.
Call $H_0:=\{h\in H: \text{order}(h)|n \}$.
$\textbf{What I have so far:}$ I propose to make a bijection
$$\Phi:\text{Hom}(G,H)\to H_0:\varphi\mapsto \Phi(\varphi)=\varphi(x_0).$$
This is well defined as for all $\varphi \in \text{Hom}(G,H):\text{order}(\varphi(x_0))|n$, we use the fact that the order of an element $x$ divides $s$ iff $x^s=e$. So then $\varphi(x_0)^n=\varphi(x_0^n)=e_H$, and then $\varphi(x_0)\in H_0$.
It is definitely an injection, since the image of a groupmorphism on a cyclic domain is determined by $\varphi(x_0)$. So, if two morphisms send $x_0$ to the same $h\in H_0$, they must be the same.
I am having trouble proving it is a surjection. Take $h\in H_0$. If there is a $\varphi \in\text{Hom}(G,H):\varphi(x_0)=h$, then all is well. But why should there be such a $\varphi$?
You can just define one as you did before. You said that a homomorphism from a cyclic group is determined by the image of the generator. Now define $\varphi_h$ by sending your generator to the given element in $h \in H_0$.