Suppose $g:X\to Y$ and $f:Y\to Z$, and $f$ is a local homeomorphism, which is to say that for any $y\in Y$ there is a neighborhood $U$ of $y$ such that $f\restriction U$ is a homeomorphism from $U$ to $f[U]\subseteq Z$, and suppose also that $h=f\circ g$ is a continuous function. Does it follow that $g$ is continuous?
If $f$ is known to be a global homeomorphism, we can argue as follows: Let $V\subseteq Y$ be open. Since $f$ is one-to-one, we have $f^{-1}[f[V]]=V$, so $h^{-1}[f[V]]=g^{-1}[f^{-1}[f[V]]]=g^{-1}[V]$. But since $f$ is an open map $f[V]$ is open and since $h$ is continuous $h^{-1}[f[V]]$ is open; thus $g^{-1}[V]$ is open and $g$ is continuous.
Weakening to local homeomorphisms, let $x\in X$ and let $V\subseteq Y$ be a neighborhood of $g(x)$; we want to find a neighborhood $W\subseteq X$ such that $g[W]\subseteq V$. By the assumption of local homeomorphism, there is an open $U\ni g(x)$ such that $f\restriction U$ is a homeomorphism, and we can assume further that $U\subseteq V$ (by intersecting with $V$ if necessary). Now it seems that a choice of $W$ most likely to work is $h^{-1}[f[U]]$. Since $f$ is an open map (a local homeomorphism is an open map), this is an open set, and it certainly contains $g(x)$. However, since $f$ is no longer injective only have $f^{-1}[f[U]]\supseteq U$ which is not good.
Edit: On further thought, I think there may be problems with the truth of the statement in the given generality. Is the theorem true if I add that $X=[0,1]\times[0,1]$, and $g(0,-):[0,1]\to Y$ and $g(-,x):[0,1]\to Y$ for each $x\in[0,1]$ are already known to be continuous (i.e. $g$ is a patchwork of continuous functions, but is not known to be jointly continuous)? I also know that $f$ is a covering map, but I'm certain that being a local homeomorphism should be enough.
The statement is incorrect. To give an counterexample, let $X = \{1, 2\}$ with the trivial topology, $Y = \{1, 2\}$ with the discrete topology, and $f: Y \to Z =\{1\}$.
Then the identity map $g: X \to Y$ is not continuous while the composition $f\circ g$ is continuous.