Looking for a Better Way to Think About Polynomial Rings

The most satisfactory (by that I mean that it seems rigorous) way I've seen of defining polynomial rings is by defining $R[X]=\bigoplus_{n=0}^\infty R$. That is, elements in $R[X]$ are of the form $(a_0,a_1, a_2, \dots)$ with all but finitely many $a_i=0$. Then one defines a product on $R[X]$ by

$(a_0,a_1,\dots, a_n, \dots)(b_0,b_1,\dots, b_n, \dots)=(a_0b_0, a_0b_1+a_1b_0, \dots , \sum_{i+j=n}a_ib_j, \dots )$.

Then after proving that this makes $R[X]$ into a ring, one can define the standard $i^{th}$ basis vector $(0, \dots, 0, 1, 0, \dots):= X^i$, with the convention that $X^0=1$.

Then $ax=xa$ means

$(a,0,0, \dots )(0,1,0, \dots)=(0,1,0,\dots)(a,0,\dots)$

which makes sense and is true; both are equal to $(0,a,0,\dots)$.

Next, for polynomials of severable variables, people usually inductively define $R[X_1, \dots , X_n] := R[X_1, \dots X_{n-1}][X_n]$, so that in particular $R[X,Y]=R[X][Y]$. Now elements in this ring are really sequences of sequences, but with the way we've defined $X^i$, we can instead just write them as sequences of things that actually look like polynomials, and then define $Y^i$ in the same way we defined $X^i$. Your question 2 is then the exact same as question one: What does it mean to say $(X,0,\dots)(0,1,0,\dots)=(0,1,0,\dots)(X,0,\dots)$?

The notation $R[x]$ may be read as "the commutative ring created by taking $R$ and adding a new element $x$." The fact that this new $x$ should commute with the existing elements is merely a part of the definition - when we're working with polynomial rings, we're generally interested in commutative rings, so we don't venture out of that territory when we add a new element.

Think of particular instances like $\mathbb Q[\sqrt[3]{2}]$. This is the set of reals generated by the rationals and $\sqrt[3]{2}$. It happens that every such rational may be written as a polynomial in $\sqrt[3]{2}$ - i.e. a sum of its powers multiplied by rational coefficients - and this follows just by distributing any expression you could come up with using rationals and $\sqrt[3]{2}$ with multiplication and addition. Of course $\sqrt[3]{2}$ commutes with every other element, because everything involved are real numbers. In a more general sense, we could consider $\mathbb Q[x]$ with $x$ standing in for some unknown complex or real number - it definitely commutes with the other elements, so everything is a polynomial in $x$ - but we can't make any reductions beyond that without knowing more about $x$. So, as far as commutative algebra is concerned, letting $x$ commute with things is the best way to represent our intuition about adding new elements to commutative rings. Abstracting a little more, we see that $R[x]$ is just the commutative ring that occurs when we stick a new element in, call it $x$, and see what happens. $x$ isn't really a variable - it's a new object we're sticking into the ring. So, if we stick both $x$ and $y$ in, given that we still want things to commute, we will end up with $xy=yx$.

That said, we certainly could consider the structure of a ring $R$ adjoined with a new element $x$ which we do not assume to commute with anything - the reason we don't is likely because structure of this is far harder to work with (each monomial has to be of the form $axbxcx\ldots$ rather than $ax^n$, though we can still always distribute into such monomials) and isn't particularly useful to algebraic number theory or Galois theory, where $R[x]$ is likely to appear.

When we write $K[X]$, the variable $X$ merely stands for the sequence $(0, 1, 0, 0, \dots)$. The product of polynomials is so defined that $X^i$ can be identified with the sequence $(0, 0, \dots, 0, 1, 0, \dots)$ with $1$ on the $i$-th position. It is natural to identify the constant $1$ with the sequence $(1, 0, 0, \dots)$. Thus, the polynomial $f = a_0 + a_1 X + \dots + a_n X^n$ can be identified with the sequence $(a_0, a_1, \dots, a_n, 0, 0, 0, \dots)$. This identification allows us to identify each polynomial with a sequence of finite support of elements from $K$. In other words, $K[X] = \{f : \Bbb N \to K \Big| \space \#|\text{supp} f| < \infty \}$.