If $f(x) = \sum_{i=0}^d a_i x^i \in \mathbb{Z}_{2^n}[x]$ is a polynomial with coefficients $a_i \in \mathbb{Z}_{2^n}$, then it is known due to Rivest that $f(x) \mod 2^n$ permutes the elements of $\mathbb{Z}_{2^n}$ if and only if $\Delta_1 = a_2 + a_4 + \ldots$ and $\Delta_2 = a_3 + a_5 + \ldots$ are even, and $a_1$ is odd. Here $\mathbb{Z}_{2^n} = \{0, 1, \ldots, 2^n -1\}$. As a result, there are permutations $F: \mathbb{Z}_{2^n} \to \mathbb{Z}_{2^n}$ that do not have a polynomial $f \in \mathbb{Z}_{2^n}[x]$ such that $F(x) = f(x) \mod 2^n$ for all $x \in \mathbb{Z}_{2^n}$. For example, consider any $F$ such that $F(0)$ and $F(1)$ are even -- see Permutations on $[2^k]$ And the Existance of Permutation Polynomials.
What confuses me is the following. Given a function $G : \mathbb{F}_{2^n} \to \mathbb{F}_{2^n}$, where $\mathbb{F}_{2^n}$ is a finite field of size $2^n$, then $G$ admits a unique univariate polynomial over $\mathbb{F}_{2^n}$ of degree at most $2^n - 1$: $G(x) = \sum_{j=0}^{2^n-1}\delta_j x^j$, where each $\delta_j \in \mathbb{F}_{2^n}$. Notice that there is always a polynomial representation for $G$ (unlike in the case above). Why does going from $\mathbb{Z}_{2^n}$ to $\mathbb{F}_{2^n}$ make such a difference, and what would the coefficients $\delta_j$ look like if they come from a finite field instead? I realize that there isn't a contradiction between both notions but I don't quite understand why, or how polynomials for $G(x)$ and $F(x)$ differ.
