False homework proof?: The image of an element has the same order.

That's correct, and that's a correct counterexample. The two things you can prove are what you said (it's true if $\phi$ is a isomorphism), or with the given condition, the order of $\phi(g)$ must divide $k$.

Injectivity is a necessary and sufficient condition for the question statement to hold, which we can see as follows. If $\phi$ is injective then the order of $g$ equals the order of $\phi(g)$ because $g^n = e_G \Leftrightarrow \phi(g)^n = e_{G'}$ so the least $n > 0$ for which $g^n = e_G$ is the least $n$ for which $\phi(g)^n = e_{G'}$. Also, if $\phi$ is not injective then there is an element $g \ne e_G$ with $\phi(g) = e_{G'}$ and then clearly the order of $g$ and the order of $\phi(g)$ differ.

Yes, you are right. All you can say in general is that the order of $\phi(g)$ divides $k$.