1. Example 2.2 (excerpt)
Example 2.2. Let $G$ be a finite group and let $k$ be a commutative ring. The group ring is the set of all functions $\alpha : G \to k$ made into a ring with pointwise operations: for all $x \in G$,
$$\alpha + \beta : x \mapsto \alpha(x) + \beta(x) \quad \text{and} \quad \alpha\beta : x \mapsto \alpha(x) \beta(x).$$
If $y \in G$, the function $\delta_y$, defined by
$$\delta_y(x) = \begin{cases} 1 & \text{if } x = y, \\ 0 & \text{if } x \neq y, \end{cases}$$
is usually denoted by $y$. It is easy to check that $kG$ is a $k$-module and that each $\gamma \in kG$ has a unique expression
$$\gamma = \sum_{y \in G} a_y y,$$
where $a_y \in k$. In this notation, elements of $G$ multiply as they do in $G$; in particular the identity element $1$ in $G$ is also the unit in $kG$. Multiplication in $kG$ is called convolution, and a formula for it is
$$\left ( \sum_x a_x x \right )\left ( \sum_y b_y y \right ) = \sum_{x,y} a_x b_y x y = \sum_z \left ( \sum_x a_x b_{x^{-1}z} \right ) z.$$
2. Questions
If $x,y \in G$ are distinct, then in $kG$ we have $xy = \delta_x \delta_y = \delta_{x,y} = 0$. I think this is somehow wrong. The example says that $1 \in G$ is also the unit in $kG$. But this contradicts my above calculation, namely $\delta_1 \delta_x = 0$ if $x \ne 1$. So I assume the multiplication formula addresses this, because you end up with terms $z \in G$ with coefficients in $k$.
I want to make sure I understand the final equality in the convolution. I'll show that both sides have the same terms. Let $a_x b_y x y$ be a term on the LHS. Taking $z = xy$, we then have the term $a_x b_y x y$ on the RHS. On the RHS, we have terms of the form $a_x b_{x^{-1} z} z$. Taking $x = x$ and $y = x^{-1}$, we also have this term on the LHS.
Why is the commutativity of $k$ required?
I think a quick concrete example should (implicitly) answer all three of your questions.
Elements of the group ring $kG$ look like polynomials with coefficients in $k$ and variables in $G$. We multiply two polynomials exactly how we always do, but at the end we have a ~bonus step~ where we additionally do multiplication in $G$. Here's an example:
Say $G = \langle \sigma \mid \sigma^2 = 1 \rangle$ is a cyclic group of order $2$ and $k = \mathbb{Q}$. Then elements of $kG$ look like $a 1 + b \sigma$, and we multiply them by
$$ (a1 + b\sigma)(c1 + d\sigma) = ac 1 1 + ad 1 \sigma + bc \sigma 1 + bd \sigma \sigma $$
of course, we can now multiply the variables using the multiplication $G$ to reduce this further to
$$ ac 1 1 + ad 1 \sigma + bc \sigma 1 + bd \sigma \sigma = (ac + bd)1 + (ad + bc) \sigma $$
This is related to the definition in Rotman because we can think of a polynoimal $\sum_g a_g g$ as the function $a_{-} : G \to k$ which takes $g$ to its coefficient $a_g$. Notice how this identifies the polynomial $g = 1_k g$ with the delta function $\delta_g$.
Now, from this discussion, do you see why $1$ (which is really an abbreviation for the polynomial $1_k 1_G$) is still the multiplicative identity? Also, related to your question (1), do you see why $(1_k g) (1_k h) = 1_k gh$, so that multiplication in $kG$ extends multiplication in $G$? It's best to ignore Rotman's comment about pointwise multiplication, since that isn't the right notion of multiplication for the group ring. (You already realized this with your computation of $xy = \delta_x \delta_y = 0$).
You're also correct in your question (2). The idea is to rewrite $y$ as $x^{-1}z$ for $z=xy$. Hopefully the example of polynomial multiplication also makes this more obvious, since our three equal terms correspond exactly to the three equal terms in Rotman's definition (and you should convince yourself of this):
$$ \left ( \sum_x a_x x \right ) \left ( \sum_y b_y y \right ) \ = \ \sum_{x,y} a_x b_y xy \ = \ \sum_z \left ( \sum_x a_x b_{x^{-1}z} \right ) z $$
$$ (a1 + b\sigma)(c1 + d\sigma) \ = \ ac 1 1 + ad 1 \sigma + bc \sigma 1 + bd \sigma \sigma \ = \ (ac + bd)1 + (ad + bc) \sigma $$
As for your question (3), we actually don't need $k$ to be commutative in order to define the group ring, and the group ring is fruitfully used in cases where $k$ isn't commutative. But as is often the case, it simplifies the theory to work over a commutative base ring, and I imagine Rotman will go on to prove some theorems that do use commutativity. In this case, making commutativity of $k$ part of the definition will keep him from having to write "let $kG$ be a group ring with $k$ commutative" all the time.
I hope this helps ^_^