How Reduction works in proving NP-Hard?
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.