Let $p:Y\to X$ be a (surjective) covering map and $G$ be the group of deck transformations of this cover. Assume $Y$ is connected. And assume, if needed, for niceness that $X$ admits a universal cover.
It is well-known that the natural projection $\pi:Y\to Y/G$ is a covering. Here is the proof. Write $G$ to denote the set of all the deck transformations of $p:Y\to X$. We will show that $$ \forall y\in Y,\exists \text{ a neighborhood } V \text{ of } y \text{ in } Y \text{ such that } V\cap gV\neq \emptyset \text{ if and only if } g \text{ is identity. } $$ This is enough to show that $Y\to Y/G$ is a covering. To do this, let $x=p(y)$ and $U$ be an evenly covered neighborhood of $x$ in $X$. Let $V$ be the slice of $p^{-1}(U)$ contaning $y$. Suppose $V\cap gV\neq\emptyset$ for some $g\in G$. Note that by definition of a deck transformation it follows that $gV$ is also a slice of $p^{-1}(U)$. Two slices intersect only if they are same. Thus $gV=V$ and so $g$ must fix $y$. But any deck transformation of a connected space which fixes a point is the identity.
But the map $p$ factors through $\pi$ to induce a map $\bar p:Y/G\to X$.
Is this map also a covering?