I am confused about the polynomials and polynomial functions. I know when we say $R[x]$, we are refering to polynomials, so what exactly is the difference? Is it basically that one is a function and the other one isn't?
There's this section of Artin's book that discusses a map from polynomials to its associated polynomial function. The first line says that the existence of the homomorphism follows from the substitution principle. What does this mean? How does it follow?
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.
Let me first be very general and consider any integral domain $R$.
For any $a \in R^n$ we have the evaluation at $a$: $$ev_a \colon R[x_1,...,x_n] \to R, \;\;\; f \mapsto f(a).$$ What you do is plugging in $a$ instead of $x = (x_1,...,x_n)$ and then calculate the result, so you get an element $f(a) \in R$ ("substituting").
Hence, the collection $(ev_a)_{a \in R}$ defines the homomorphism $\varphi \colon R[x_1,...,x_n] \to \mathfrak R$, where $\mathfrak R$ is the ring of polynomial functions on $R$.
You also asked about the difference of polynomials and polynomial functions. Of course they are not "the same" in a strict sense, but one can wonder if we can identify them via $\varphi$. Suppose that $R$ is a finite ring. Then there are infinitely many polynomials with coefficients in $R$: $$X, X^2, X^3, X^4, ... \in R[X]$$ are all different polynomials. However, there are only $|R|^{|R|}$ functions $R \to R$, so polynomials and polynomial functions cannot be the same. In other words, $\varphi$ cannot be injective! Hence, a necessary condition for $\varphi$ to be injective is that $R$ is infinite.
In fact, it is sufficient:
Lemma: Suppose that $R$ is an infinite integral domain. Then for any $n$, the map $\varphi \colon R[x_1,...,x_n] \to R$ is injective.
The proof is not really difficult. You should try it (use induction on $n$).
