Projection between graphs extends to a covering space This is exercise 1.A.10 on page 87 of Hatcher's book Algebraic topology.

Let $X$ be the wedge sum of $n$ circles, with its natural graph structure,
  and let $\tilde X \to X$ be a covering space with $Y \subset \tilde X$ a finite connected subgraph. 
Show there is a finite graph $Z ⊃ Y$ having the same vertices as $Y$, such that
  the projection $Y→X$ extends to a covering space $Z→X$.

Question:
How can we obtain graph $Z$ by adding edges from the given finite graph $Y$?
This turned out to be Marshall Hall's Theorem in Stallings' article  Topology of finite graphs. But I need more explanations.
For example, an answer says for the case where $X=S^1 \vee S^1$, $\tilde X$ is the universal covering space, and $Y=B(1,n)$, i.e. the ball of radius $n$ and centered at $1$ in the graph $\tilde X$, the covering space $Z \to X$ extending projection $Y \to X$ is just $B(1,n)$ with additional edges between vertices of the sphere $S(1,n)$.
Could someone give an explanation for this?

Edit 1:
This can be used to prove the residual finiteness of finitely generated free group, which states:

If $F$ is a finitely generated free group and $x \in F$ is not the identity element, then there's normal subgroup $H \subset F$ of finite index s.t. $x \not \in F$, hence $x$ has nontrivial image in a finite quotient group of $F$.

Edit 2:
This can be used to prove this:

Let $F$ be a finitely generated free group and $H$ be a finitely generated subgroup of $F$. Let $x∈F−H$. Then there is a finite index subgroup $K$ of $F$ such that $H⊆K$ and $x∉K$.

 A: Maybe I'm just a graph theorist who doesn't see why the topologists are making a big deal out of all this...
But I'm pretty sure that a covering space of $X$ is equivalent to a (possibly infinite) directed multigraph whose edges are $n$-colored in such a way that every vertex is incident to exactly one incoming and one outgoing edge of each color. 
The finite connected subgraph $Y$ is therefore a directed multigraph whose edges are $n$-colored in the same way - but with "exactly one" replaced by "at most one".
The claim in Hatcher's exercise is just that we can patch the "at most" problem and get back to exactly one incoming and outgoing edge of each color, by adding only edges and not vertices.
To see this, let's deal with one color at a time. There are three kinds of problem vertices for the $i^{\text{th}}$ color:


*

*Vertices with an incoming edge but not an outgoing edge of the $i^{\text{th}}$ color.

*Vertices with an outgoing edge but not an incoming edge of the $i^{\text{th}}$ color.

*Vertices with neither of those.


Because the total number of incoming edges of the $i^{\text{th}}$ color is the same as the total number of outgoing edges, and everything is finite, there is an equal number of vertices of type 1 and type 2. So we can patch those vertices by pairing them up, and adding an edge of the $i^{\text{th}}$ color from each type 1 vertex to the paired type 2 vertex.
To patch type 3 vertices, just add a loop of the $i^{\text{th}}$ color at those vertices.
