Let $k$ be a field and $V\subseteq k^n$ be an affine variety. We identify the coordinate ring $k[V]$ as the quotient ring $k[x_1,\cdots ,x_n]/I(V)$. In a lot of texts on algebraic geometry, when $V=k^n$, then the coordinate ring $k[k^n]$ is identified as the polynomial ring $k[x_1,\cdots ,x_n]$
This definition makes sense when the field is infinite. In that case, $I(V)=\{0\}$, so every equivalence class in $k[x_1,\cdots ,x_n]/I(V)$ has exactly one element, hence $$k[k^n]\cong k[x_1,\cdots ,x_n]/\{0\}\cong k[x_1,\cdots ,x_n]$$ But what if $k$ is finite? Take $n=1$ and $k=\mathbb{F}_2$. Consider the equivalence class $[0]$ in $\mathbb{F}_2[x]/I(\mathbb{F}_2)$. Since the functions $f=x^2+x$ and $g=x^4+x^3+x^2+x$ (and more) vanish on $\mathbb{F}_2$, we must have the equivalence class $[0]$ has more than one element (in fact, infinitely many)
In this case, is it still appropriate to identify the coordinate ring to the polynomial ring? I asked this because none of the texts online explain this ambiguity of the identification
Polynomials in $n$ variables over a finite field $F$ just cannot be identified with functions on $F^n$, for exactly the reason you describe. What they can be identified with is a little more complicated. Over any field $F$ you can think of a polynomial in $n$ variables over $F$ as a function, not only on $F^n$, but on $L^n$ for any field extension $L$ of $F$, and in fact on $A^n$ for any $F$-algebra $A$, and this separates points in the sense that two polynomials define the same collection of such functions iff they are identical as polynomials (meaning they have the same monomials with the same coefficients).