How Reduction works in proving NP-Hard? A problem $X$ is $NP$-Hard if for all $Y \in NP$, $Y \leq_P  X$. Further, if a problem $Z$ is $NP$-Complete, and $Z \leq_P  X$, then I can prove (rather mechanically) that $X$ is $NP$-Hard.

I also found that, for a given X, say TSP, to prove it is $NP$-Hard, we often find a ($NP$-Complete) Z say $HAM-CYCLE$ (Hamiltonian-Cycle), and try to show that $HAM-CYCLE$ can be reduced to a $TSP$ in polynomial time. 

My confusion is, why don't we try to show that for each instance of $TSP$, there exists a corresponding instance of $HAM-CYCLE$. Specifically, what if there exists an instance of $TSP$, for which there is no corresponding instance in $HAM-CYCLE$! In this case, how can the prior knowledge about the hardness of $HAM-CYCLE$ help in inferring on TSP's hardness!

Note: I also had similar concerns with proving $NP$-complete class. However, since all $NPC$ are also $NP$, I felt, similar reduction of a known NPC problem to a given problem, say Q, works. However, for $NP$-hard case, a given problem X need not to be $NP$ at all.
Apparently, you are expecting the reduction between any two $\mathsf{NP}$-complete problems $X$ and $Y$ to be a one-to-one map (e.g., a bijection). The idea of a reduction, however, is irrelevant to having instances "correspond" to others. What reductions do give us is a hardness relation between problems (and not between individual instances). If $A$ and $B$ are problems and $A$ is reducible to $B$, then we expect $B$ to be harder than $A$; the idea is that, if there is an efficient algorithm which solves $B$, then we can also solve $A$ by reducing it to $B$ and running said algorithm. Completeness, then, corresponds to the notion of hardness equivalence: $A$ is reducible to $B$ and $B$ is reducible to $A$; that is, if $A$ is hard (resp., easy), then so is $B$, and vice-versa.

To give an example why we cannot generally expect reductions to be injective, surjective, or even bijective, let again $X$ to be $\mathsf{NP}$-complete and consider the problem $Z = \{ (0, x) \mid x \in X \} \cup \{ (1, y) \mid y \in \{ 0,1 \}^\ast \}$. Naturally, we can reduce $X$ to $Z$ by mapping each instance $x$ of $X$ to $(0,x)$; this does not give us a surjective map. Further, note $Z$ is also reducible to $X$: map $(0,x)$ to $x$ and $(1,y)$ to $x'$ where $x' \in X$ is a fixed (i.e., hard-coded) instance. This reduction is surjective, though not injective. As you can see, there is really no relation between what we expect from a reduction and injectivity or surjectivity.