My text (p. 30, eq. 2.17) states the "commutative law" for sums as
$$\sum\limits_{k\in K}{a_k} = \sum\limits_{p(k)\in K}{a_{p(k)}}$$
where $p$ is a permutation and $K$ is a set of integers.
While I understand how this rule, as stated, can be applied, I don't understand how this can be a complete statement of commutativity, or suffice for the manipulations made subsequently (in fact, immediately following; second equality of second to last equation on same page) in the text.
Isn't it more complete (and necessary of the law is to be used as it seems to be in subsequent discussion) to say
$$\sum\limits_{k\in K}{a_k} = \sum\limits_{k\in p(K)}{a_{q(k)}}$$
where $K$ is ordered, $p(K)$ is $p$ applied to each element of $K$, and $p$ and $q$ are any permutations (including $p=q$ and $p(k)=q(k)=k$)?
As stated in the text, saying $p(k)\in K$ is no different from saying $k\in K$ (since $K$ has no order), an failing to distinguish $p$ and $q$ misses the opportunity to independently rearrange summand expression and the summation order (a technique used often in the text).
It’s probably better to think of this commutative law as a renaming law: it says that as long as the renaming is bijective and is carried out consistently, it doesn’t affect the value of the summation. The consistency is enforced by having $p(k)$ both in the range of summation and in the summand. And the consistency is important. Suppose that $K=\{1,2\}$, $a_k=k$, $p:\Bbb Z\to\Bbb Z:k\mapsto 3-k$, and $q:\Bbb Z\to\Bbb Z:k\mapsto k+3$. Then
$$\sum_{p(k)\in K}a_{q(k)}=\sum_{3-k\in K}(k+3)=9\ne3=\sum_{k\in K}k=\sum_{k\in K}a_k\;.$$
Let’s take a look at the specific example that you mention,
$$\sum_{0\le n-k\le n}\big(a+b(n-k)\big)=\sum_{0\le k\le n}(a+bn-bk)\;,$$
to see that this renaming law actually suffices.
Let $K=\{0,1,\ldots,n\}$ and $p:\Bbb Z\to\Bbb Z:k\mapsto n-k$; $p$ is a permutation of the integers, and $p[K]=K$. Then
$$\begin{align*} \sum_{0\le n-k\le n}\big(a+b(n-k)\big)&=\sum_{p(k)\in K}\big(a+bp(k)\big)\\ &=\sum_{k\in K}(a+bk)&(\text{commutativity})\\ &=\sum_{p(k)\in p[K]}(a+bk)&(p\text{ is a permutation})\\ &=\sum_{p(k)\in K}(a+bk)&(p[K]=K)\\ &=\sum_{p(k)\in K}\big(a+bn-b(n-k)\big)\\ &=\sum_{p(k)\in K}\big(a+bn-bp(k)\big)\\ &=\sum_{k\in K}(a+bn-bk)&(\text{commutativity})\\ &=\sum_{0\le k\le n}(a+bn-bk)\;. \end{align*}$$