Map from polynomial to polynomial function

It may help to think of the polynomial ring $R[X]$ differently, at least for a short moment, as the group $\bigoplus_{n\in\Bbb{N}}R$ with multiplication defined by $$(c_k)_{k\in\Bbb{N}}\cdot(d_k)_{k\in\Bbb{N}}=\left(\sum_{j=0}^kc_jd_{k-j}\right)_{k\in\Bbb{N}}.$$ Here of course the sequences $(c_k)_{k\in\Bbb{N}}$ and $(d_k)_{k\in\Bbb{N}}$ correspond to the polynomials $$\sum_{k\in\Bbb{N}}c_kX^k \qquad\text{ and }\qquad \sum_{k\in\Bbb{N}}d_kX^k,$$ where the sums are finite by definition of the direct product. The newly defined product of the sequences above then indeed corresponds to the product of these polynomials, so this ring is isomorphic to $R[X]$. In this ring the powers of the indeterminate $X$ correspond to the standard basis elements of the direct sum. They are in no way functions from $R$ to $R$.

Now every such sequence does define a function $R\ \longrightarrow\ R$ by substitution, i.e. by plugging in the elements of $R$. In this way the sequence $(c_k)_{k\in\Bbb{N}}\in R[X]$ defines the function $$R\ \longrightarrow\ R:\ r\ \longmapsto\ \sum_{k\in\Bbb{N}}c_kr^k.$$

Of course the same ideas work for the polynomial ring in $n$ indeterminates, by repeating this construction $n$ times.

To give an example for the difference between polynomials and polynomial functions the following might help:

Take any prime $p \in \mathbb{Z}$ and look at $\varphi: \mathbb{Z}/p\mathbb{Z} \longrightarrow \mathbb{Z}/p\mathbb{Z}$ with $\varphi(x)=x^p-x$. This is a function, not a polynomial. Two functions are equal iff their domain, their target and all of their values are the same. So here $\varphi$ is the same function as the zero function, because of Fermat ($x^p=x$).

But if we look at $x^p-x$ and $0$ as polynomials, namely as elements of $\mathbb{Z}/p\mathbb{Z}[X]$, they are not equal since $x^p-x$ has nonzero coefficients.

To summarize this means that different polynomials can give the same polynomial function, so we can not think of them as the same objects.