Let $x\in X$ and $V$ be an open neighborhood of $x$. There is an evenly covered open neighborhood $U$ of $p(x)$. Say $x\in U_\alpha$ and $p_\alpha:U_\alpha\to U$ is the restriction of $p$ and a homeomorphism. Then $V\cap U_\alpha$ is homeomorphic to $p(V\cap U_\alpha)$. This is an open set containing $p(x)$, so it contains a path-connected neighborhood $W$. You can then just take the preimage of $W$ under $p_\alpha$.
Remark: The local (path-)connectedness of $X$ makes $Y$ local (path-)connected, too, which is very easy to prove since $p$ is continuous and open. But it can be shown that even a quotient map carries over the local (path-)connectedness from the domain to the image. (This is not difficult either, if you use the right characterization of local connectedness.)
Let $p: \tilde{X} \rightarrow X$ be a covering map in the broad definition (empty sum allowed, as Hatcher does). First, $p[\tilde{X}]$ is open in $X$: let $y \in X$ with $x \in \tilde{X}$ such that $p(x) = y$. Let $U$ be an evenly covered neighbourhood of $y$, so $p^{-1}[U] = \sum_{i \in I} O_i$ (disjoint sum over index set $I$). As $x$ is in the sum, it is non-empty: some $O_{i_{0}}$ contains $x$ and $p |_{O_{i_0}}$ is a homeomorphism between $O_{i_{0}}$ and $U$. In particular $U \subset p[\tilde{X}]$, so $y$ is an interior point of $p[\tilde{X}]$.
Suppose $y \notin p[\tilde{X}]$. Then again take an evenly covered neighbourhood $U$ of $y$: $p^{-1}[U] = \sum_{i \in I} O_i$, disjoint sum over some index set $I$, where for every $i \in I$, the map $p|_{O_i}$ is a homeomorphism between $O_i$ and $U$. This means (!) that $I = \emptyset$, as otherwise we'd have a preimage for $y$, contradicting how we picked $y$. So in fact $p^{-1}[U] = \emptyset$, and this shows that $U \subset X\setminus p[\tilde{X}]$, so $p[\tilde{X}]$ is closed.
Now if $X$ is connected and $\tilde{X}$ is non-empty, then $p[\tilde{X}]$ is closed, open and non-empty, so equals $X$ by connectedness. So $p$ is surjective.
These are the only conditions we need: $X$ connected and $\tilde{X} \neq \emptyset$.
Best Answer
I think a cover of a cover of the Hawaiian earring gives an example where the composition fails to be a covering space, and the space $X$ is path conencted
See Exericse 6 on page 79 of Hatcher's book
Edit: The composition will be a covering map if the fiber $q^{-1}(z)$ is finite (proof) or, equivalently, the space $Z$ is semi-locally simply connected (In particular the Hawaiian earring is not semi-locally simply connected).