Ginger Rogers did everything Fred Astaire did, only backwards and in high heels.
I shall promote my comment regarding the first part to a full answer. As I said, consider
$$\sum_{0\leq k\leq n}(a+bk)=(a+0\cdot b)+(a+1\cdot b)+\cdots+(a+(n-1)\cdot b)+(a+n\cdot b)$$
and contrast with
$$\sum_{0\leq k\leq n}(a+(n-k)b)=(a+(n-0)\cdot b)+(a+(n-1)\cdot b)+\cdots+(a+(n-(n-1))\cdot b)+(a+(n-n)\cdot b)$$
Writing the sums in this way shows that one is merely a reversal of the other, and this indeed is an application of addition's commutativity. (The substitutions done by Marvis and Brian say the same thing, only in algebraic language.)
Since you've said that you're working on Concrete Mathematics, I feel even more justified to show the utility of Iverson brackets in manipulating double sums. (Effectively, this is a restatement of Marvis's and Brian's answers in Iversonian form.)
As a reminder of the definition, $[p]$ evaluates to $1$ if condition $p$ is true, and $0$ if condition $p$ is false. Iverson brackets have the useful property that $[p\text{ and }q]=[p][q]$, which will be important here.
We can rewrite your double sum in Iversonian form as
$$\sum_j\sum_k\frac{[1\leq j<k\leq n]}{k-j}$$
(It's a bit of an abuse that only one summation sign was written when two indices are indicated, but let's forgive that.) Note that we can effectively treat the sum as an infinite summation, since the Iverson brackets zero out any terms that do not satisfy the condition it encloses.
As already explained in the previous answers, we can do the substitution $k\mapsto k+j$ like so:
$$\begin{align*}\sum_j\sum_k\frac{[1\leq j<k+j\leq n]}{k+j-j}&=\sum_j\sum_k\frac{[1\leq j<k+j\leq n]}{k}\\&=\sum_j\sum_k\frac{[1\leq k<k+j\leq n]}{k}\end{align*}$$
Note that here it was fine not to touch the summation's indices, since if $-\infty < j < \infty$ and $-\infty < k < \infty$, we also have $-\infty < j+k < \infty$. We are also justified in replacing the $j$ in the second line with a $k$, since $k < k+j$ in the range being considered
We can now split the Iverson factor using the property I mentioned earlier. In particular, we have
$$\begin{align*}\sum_j\sum_k\frac{[1\leq k<k+j\leq n]}{k}&=\sum_k\sum_j\frac{[(1\leq k\leq n)\text{ and }(1\leq j+k\leq n)]}{k}\\&=\sum_k\sum_j\frac{[1\leq k\leq n][1\leq j+k\leq n]}{k}\end{align*}$$
(I've also taken the liberty of swapping summation order in the meantime.)
We can rearrange the last bit to
$$\sum_k\sum_j\frac{[1\leq k\leq n][-k\leq j\leq n-k]}{k}=\sum_k\sum_j\frac{[1\leq k\leq n][1\leq j\leq n-k]}{k}$$
How did this happen? Due to the conditions $1\leq j\leq n$ and $1\leq k\leq n$, we can zero out all the terms from $j=-k$ to $j=0$. A final rearrangement leads to
$$\sum_k [1\leq k\leq n] \sum_j [1\leq j\leq n-k] \frac1{k}=\sum_{1\leq k\leq n} \sum_{1\leq j\leq n-k} \frac1{k}$$
and we are done.
Best Answer
As Derek already said, there is no essential difference between functions $A\times B \to C$ and functions $A\to (B \to C)$ via Currying (this is also more abstractly expressed by the universal property of an exponential which unifies the set-theoretical currying and currying in a typed lambda calculus).
On the notational side of things, I personally prefer $x\mapsto f(x)$ to $\lambda x. f(x)$ and I suspect many other mathematicians feel the same (especially since $\lambda$ is such a commonly used letter).
EDIT: (now that my answer stopped being one, let me add some rambling that the 29 people so far have not upvoted for):
I'm guessing many mathematicians are less "comfortable" with nested expressions like $v\mapsto (f \mapsto f(v))$. That would be nothing extraordinary, since there are various concepts that some mathematicians feel less comfortable about. Here are two (unrelated) things that I have encountered:
Although, your example:
is fine and not hard to understand, in my opinion. For every $v\in V$ we have $Av\in V^{**}$, i.e. $Av : V^* \to \mathbb K$. Hence we can plug in an $f\in V^*$ to get $f(v) \in \mathbb{K}$. If the author thinks it is easy to understand and is more used to it than $v\mapsto (f \mapsto f(v))$ then they would obviously have no reason to change the notation.
So the reason why $v\mapsto (f\mapsto f(v))$ (or a variant thereof) is not used as much is probably: "I'm not used to this notation and I'm perfectly happy with mine."
By the way, my personal favourite is also not: $$A : V \to V^{**}, v \mapsto (f\mapsto f(v))$$ but $$A : V\to V^{**}, v\mapsto \_(v)$$ where it is implied that $\_$ is a placeholder, i.e. $\_(v) : V^* \to \mathbb{K}, f\to f(v)$.