How to prove that R[x] is commutative

You need to prove that polynomials satisfy basic axioms of ring for addition and multiplication.

Let $P(x)=\sum_{i=0}^na_ix^i, \:Q(x)=\sum_{i=0}^mb_ix^i$.Then $$ P(x)+Q(x)=\sum_{i=0}^n(a_i+b_i)x^i=\sum_{i=0}^n(b_i+a_i)x^i=Q(x)+P(x) $$ This proves polynomials is commutative for addition. $$ P(x)Q(x)=\sum_{i=0}^n\sum_{j=0}^i(a_jb_{i-j})x^i=\sum_{i=0}^n\sum_{j=0}^i(b_{i-j}a_j)x^i=Q(x)P(x) $$ This proves polynomials is commutative for multiplication.

Let $-P(x)=\sum_{i=0}^n(-a_i)x^i$. Then $$ P(x)+(-P(x))=\sum_{i=0}^n(a_i+(-a_i))x^i=0 $$ So $-P(x)$ is the negative element or additive inverse of $P(x)$.

Let $1(x)=1$. Then $$ P(x)1(x)=\sum_{i=0}^n(a_i\cdot 1)x^i=P(x) $$ So $1(x)$ is the unity of polynomials ring.

Similarly you can prove the associativity and distributivity for addition and multiplication.

The set of polynomials $R[x]$ is the collection of formal composites of the form $\sum_{i=0}^na_ix^i$ with $a_i\in R$. In practice what this means is that a polynomial is identified with its sequence of coefficients $(a_i)_{i\in\Bbb N}$, which is finitely supported (that is, there is some $n$ such that $a_i=0$ for $i>n$). (If we drop the finite support restriction, we get the ring of power series $R[[x]]$ instead.) In keeping with traditional notation, we will write $\sum_ia_ix^i$ instead of $(a_i)_{i\in\Bbb N}$, with the understanding that the sum and $x$ are just (suggestive) notation. Once we have proven the ring properties, we can show that there actually is an element $x=\sum_i\left(\begin{cases}1&i=1\\0&i\ne1\end{cases}\right)x^i$, and then we can actually take a power $x^i$ and multiply it with $a_i$ and sum to reconstruct the element $a=\sum_ia_ix^i$, thus justifying the notation.

The addition of two coefficient functions is defined by $\sum_i(a+b)_ix^i=\sum_i(a_i+b_i)x^i$. From this it is simple to verify the group laws for addition, where the identity is $0=\sum_i0x^i$ and a negative is $-\sum_ia_ix^i=\sum_i(-a_i)x^i$.

The multiplication of two coefficient functions is defined by $\sum_i(ab)_ix^i=\sum_i(\sum_{j=0}^ia_jb_{i-j})x^i$. (Here the inner summation over $j$ is a "real" summation, taken within the ring.) This is closed in $R[x]$ because $\sum_{j=0}^ia_jb_{i-j}\in R$. To verify the ring properties, we need to show, for example, that $\sum_i(a(b+c))_ix^i=\sum_i(\sum_{j=0}^ia_j(b_{i-j}+c_{i-j}))x^i$ is equal to $\sum_i(ab+ac)_ix^i=\sum_i(\sum_{j=0}^ia_jb_{i-j}+\sum_{j=0}^ia_jc_{i-j})x^i$, which amounts to proving that $$\sum_{j=0}^ia_j(b_{i-j}+c_{i-j})=\sum_{j=0}^ia_jb_{i-j}+\sum_{j=0}^ia_jc_{i-j}$$ for $a_i,b_i,c_i\in R$, which follows from the ring axioms for $R$. Don't let anyone tell you these are trivial - sometimes notation which is too suggestive can make these things seem obvious, but there really is something to prove. (Associativity requires a complicated diagonal commutation over the sum indexes in a triangular region, which is not at all obvious.)

But let's move on to the real question, commutativity. We want to show that $\sum_i(ab)_ix^i=\sum_i(\sum_{j=0}^ia_jb_{i-j})x^i$ is equal to $\sum_i(ba)_ix^i=\sum_i(\sum_{j=0}^ib_ja_{i-j})x^i$. The key step here is that sums can be reversed - we can substitute $k=i-j$ for the index of the summation to get $$\sum_{j=0}^ia_jb_{i-j}=\sum_{k=0}^ia_{i-k}b_k=\sum_{k=0}^ib_ka_{i-k}=\sum_{j=0}^ib_ia_{i-j},$$ as we wanted to show.