Why does there exist a deck transformation mapping here? See Kevin Dong's answer here.

Let $\widetilde{\gamma}$ be a path in $X_N$ based at $x_0^N \in X_N$ with $p(\widetilde{\gamma}) = \gamma$ a nontrivial path in $X$. Since the cover is normal, there exists a deck transformation mapping $x_0^N$ to any other preimage under $p$ of $x_0 \in X$. (Why?)

I do not follow why the existence of such a deck transformation mapping follows from the normality of the cover... can someone explain?

From Wikipedia:

Now suppose $p:C \to X$ is a covering map and $C$ (and therefore also $X$) is connected and locally path connected. The action of $Aut(p)$ on each fiber is free. If this action is transitive on some fiber, then it is transitive on all fibers, and we call the cover regular (or normal or Galois).

By the very definition of what a normal cover is, the fact that the cover $p : X_N \to X$ is normal implies that for any $x_0^N \in p^{-1}(x_0)$ and any other point $y \in p^{-1}(x_0)$, there is a deck transformation $\sigma \in Aut(p)$ such that $\sigma(x_0^N) = y$. This is what an action being transitive means. Here, the cover is normal because it corresponds to a normal subgroup of $\pi_1(X)$, a standard fact in algebraic topology (you can find the proof in any textbook).
