Your comment is close to the mark, but to get exactly what you want takes a bit of work.
The point is that the map $\phi$ can be extended to a covering map, by enlarging the graph $\Gamma$. To be precise:
Lemma: There exists a graph $\Delta$, an embedding $h : \Gamma \to \Delta$ which is a homotopy equivalence, and a covering map $q : \Delta \to \Gamma'$, such that $q \circ h = \phi : \Gamma \to \Gamma'$.
Once that has been proved, then you can apply the fundamental group functor (assuming a chosen base point in $\Gamma$, and the image base points in $\Delta$ and $\Gamma'$):
- $h_* : \pi_1(\Gamma) \to \pi_1(\Delta)$ is an isomorphism;
- $q_* : \pi_1(\Delta) \to \pi_1(\Gamma')$ is an injection;
- $q_* \circ h_* = \phi_* : \pi_1(\Gamma) \to \pi_1(\Gamma')$.
Then, from your assumption that $\phi_*$ is a surjection, it follows that $q_*$ is a surjection hence a group isomorphism. From that it follows that the covering map $h$ is a graph isomorphism, from which it follows that $\phi$ is surjective and hence a homeomorphism.
To prove the lemma, consider a vertex $v \in \Gamma$ and $w=\phi(v) \in \Gamma'$. To say that $\phi$ is a local covering map at $v$ means that there exists a neighborhood $U$ of $v$ in $\Gamma$ such that the restriction $\phi : U \to \Gamma'$ is a homeomorphism onto a neighborhood $\phi(U)$ of $w$ in $\Gamma'$. If $\phi$ is a local covering at every vertex of $\Gamma$ then $\phi$ is a covering map and the lemma is proved.
Let's saythat a vertex $v \in \Gamma$ is problematic if $\phi$ is not a local covering near $v$. What do we with a problematic vertex $v \in \Gamma$? Very roughly speaking, we attach to each problematic vertex $v \in \Gamma$ a certain tree that I'll denote $T_v$, in such a way that the map $\phi$ extends over all of these attached trees to obtain the desired covering map $q$. Because we've just attached a bunch of trees, the inclusion of $\Gamma$ into the extended graph is a homotopy equivalence.
So now I'll describe these trees $T_v$, how they attach to their points $v$, and how to extend $\phi$. It will turn out that $T_v$ is a certain subtree of the universal covering space of $\Gamma'$.
Let $N_v(\Gamma) \subset \Gamma$ be the regular heighborhood of $v$, meaning the union of those oriented edges of the 2nd barycentric subdivision of $\Gamma$ which have initial vertex $v$. Because $\phi$ is an immersion, it restricts to an embedding $D_v \phi : N_v(\Gamma) \to N_w(\Gamma')$. The property that $v$ is problematic is equivalent to the property that the embedding $D_v \phi$ is not a graph isomorphism.
Let $q : \tilde\Gamma' \to \Gamma'$ be a universal covering map. Choose $\tilde w \in \widetilde \Gamma'$ such that $p(\tilde w)=w$. The map $q$ restricts to a graph isomorphism $D_{\tilde w} q : N_{\tilde w} \widetilde\Gamma' \to N_w \Gamma'$. By composition we obtain an embedding
$$(D_{\tilde w}q) ^{-1} \circ D_v \phi : N_v\Gamma \to N_{\tilde w} \widetilde \Gamma
$$
Define $T_v$ to be the closure of the union of those connected components of $\widetilde\Gamma' - \{\tilde w\}$ that are disjoint from the image of the embedding $(D_{\tilde w}q) ^{-1} \circ D_v \phi$. Note that $v$ is problematic if and only if $T_v$ is nonempty, in which case the closure is obtained by just adding the point $\tilde w$ to that union of connected components.
Take the disjoint union of all of the trees $T_v$, one for each problematic vertex $v \in \Gamma$, and attach each $T_v$ to its vertex $v$ by identifying $\tilde w$ with $v$. Extend $\phi$ over each tree $T_v$ by defining the extension to be the restriction $q \mid T_v$ of the universal covering map $q : \widetilde \Gamma' \to \Gamma'$.
That's it; that's the construction which proves the lemma. There's still work to do to prove that this construction says what the lemma says it says. But rather than going through that work, let me say intuitively what just happened.
At each problematic vertex $v$, there are a bunch of directions at $w=\phi(v)$ which are missing from the image of $D_v \phi$. To fix this problem, one then attaches a bunch of new edges to $v$. But now, the opposite endpoints of these new edges are a bunch of new problematic vertices. To fix this problem, one attaches another bunch of new edges. But now the opposite endpoints of those new edges are a bunch more of new problematic vertices. Et cetera, et cetera. To formalize all of this, do what I wrote above, and the set of all edges one adds by starting from $v$ and continuing (et cetera et cetera) is precisely the tree $T_v$.
