I'll use the definition of covering map as it appears in Hatcher's Algebraic Topology: A continuous map $p:Y\to X$ is called a covering map, if for every $x\in X$ there is an open neighborhood $U$ around $x$ whose preimage is a (possibly empty) disjoint union of open sets, each of which is mapped homeomorphically onto $U$ via $p$. Note that this definition does not require a covering map to be surjective.
Still, if the codomain is connected, the map must be surjective by the following argument:
Assume $x\notin p(Y)$. Then its preimage $p^{-1}(x)$ is empty. There is an open $U$ containing $x$ such that $p^{-1}(U)$ equals $\bigsqcup_{\alpha\in I}U_\alpha$ where $U_\alpha\approx U$ and all $U_\alpha$ are open. But since $x$ is not in the image of $p$, the disjoint union must be the empty union. This means that $U$ does not intersect $p(Y)$, so $p(Y)$ is closed.
On the other hand, $p$ is an open map. If $V\subset Y$ is an open set containing $y$ and $U$ is the evenly covered neighborhood of $p(y)$, then $y$ is contained in some $U_\alpha$. Since $V\cap U_\alpha$ is open, $p(V\cap U_\alpha)$ is an open subset of $U$, thus an open set in $X$ which is contained in $p(V)$, so $p(V)$ is open.
If $X$ is connected, then $p(Y)$ must be all of $X$, being a clopen subset of a connected space.
To prove that the restriction of the $p$ in your problem to the connected component $Z$ is also a covering map, take an $x\in X$ and an open neighborhood $U$ such that $p^{-1}(U)$ equals $\bigsqcup_{\alpha\in I}U_\alpha$ where $U_\alpha\stackrel p\approx U$ and all $U_\alpha$ are open. Since $X$ is locally connected, there is an open connected $V$ such that $x\in V\subset U$. Its preimage is $\bigsqcup_{\alpha\in I}V_\alpha$ where $V_\alpha$ is simply the preimage of $V$ in the particular $U_\alpha$. Each $V_\alpha$ is connected, so it is either entirely in $Z$ or it is disjoint to $Z$. If you delete all the $V_\alpha$'s which are not in $Z$ from the union, you obtain the preimage of $V$ under the restriction of $p$. This means that this restricted $p$ still is a covering map.
The converse does not hold. If conditions 1 and 2 hold, then $C$ is the connected component of $p$, but it is not necessarily the case that every point outside the component can be separated from $p$ by open sets.
To see that 1 and 2 imply that $C$ is the connected component of $p$, let $K$ be the connected component of $p$ in $X$. We need to show that $K = C$. Since by definition connected components are maximal connected sets, and $C \cup K$ is a connected set (the union of two connected sets with nonempty intersection is connected) containing $K$, it follows that $C \cup K = K$, i.e. $C \subset K$. For the revers inclusion, consider $q \in X \setminus C$. By assumption 2, there are disjoint open $U,V$ such that $p \in U$, $q\in V$, and $X = U \cup V$. Then $(U \cap K)$ and $(V \cap K)$ are disjoint relatively open (in $K$) sets with $K = (U \cap K) \cup (V \cap K)$, and by the connectedness of $K$ it follows that $U \cap K = \varnothing$ or $V \cap K = \varnothing$. Since $p \in U \cap K$ we must have $V \cap K = \varnothing$, in particular $q \notin K$. Since this holds for all $q \in X \setminus C$, the inclusion $K \subset C$, and consequently $K = C$, is proved.
To see that the converse does not hold, an example suffices. Let
$$X = \{(0,0),(0,1)\} \cup \bigcup_{n = 1}^{\infty} \bigl\{ (1/n,t) : 0 < t < 1\bigr\}$$
with the subspace topology inherited from $\mathbb{R}^2$. Then the connected component of $p = (0,0)$ in $X$ is the singleton $\{(0,0)\}$, but $q = (0,1)$ cannot be separated from $p$ by open sets. The quasicomponent of $p$ is the two-element set $\{p,q\}$.
By definition, the quasicomponent of a point is the intersection of all clopen (that is, closed and open) sets containing that point. The quasicomponent of $p$ consists of exactly those points of $X$ that cannot be separated from $p$ by open sets. For if $r$ does not belong to the quasicomponent of $p$, then there is a clopen set $A$ containing $p$ but not $r$, then $U = A$ and $V = X\setminus A$ is a decomposition of $X$ into two disjoint open sets that separates $p$ from $r$. And if $r$ can be separated from $p$ by open sets $U,V$, then these two sets are actually clopen, so $r$ does not belong to the quasicomponent of $p$.
To see that in the above example $q$ belongs to the quasicomponent of $p$, consider a clopen $A$ containing $p$. Since $A$ is open, there is an $N$ such that $A \cap \{(1/n,t) : 0 < t < 1\} \neq \varnothing$ for all $n \geqslant N$. Since the segment $L_n = \{(1/n,t) : 0 < t < 1\}$ is connected (even path-connected), and $L_n \cap A,\, L_n \setminus A$ is a decomposition of $L_n$ into disjoint open sets, one of these must be empty, thus $L_n \subset A$ for all $n \geqslant N$. But $q$ is the limit of the sequence $\bigl((1/n, 1-1/n)\bigr)_{n \geqslant N + 1}$ all of whose points belong to $A$ by what we've seen, and $A$ is closed, so it follows that $q \in A$. Showing that no other point belongs to the quasicomponent of $p$ is easy: $X \setminus L_n$ is a clopen set containing $p$ but none of the points on $L_n$.
And finally, since $\{p,q\}$ is not connected, it follows that the component of $p$ is a proper subset of its quasicomponent.
Best Answer
A connected component of a space $Y$ is a maximal connected subspace $C \subset Y$. This means
As you say, the connected component of $p \in Y$ is the union of all connected subsets of $Y$ containing $p$, thus it is a connected component of $Y$ by definition.
Formally you can introduce an equivalence relation on $Y$ by $p \sim q$ if there exists a connected $C \subset Y$ containing both $p, q$. The only non-trivial part here is the transitivity of $\sim$. This follows from the fact that the union of connected subspaces having non-empty intersection is connected.
The equivalence classes with respect to $\sim$ are called connected components. It is easy to see that they are precisely the maximal connected subspaces as defined above. Clearly the equivalence class of a point $p \in Y$ is the union of all connected subsets of $Y$ containing $p$.