Probably a change of variables may be of help to make things clear.
For instance, in the summation
$$\sum_{0 \leq k \leq n}(a+bk)=\sum_{0 \leq n-k \leq n}(a+b(n-k))=\sum_{0 \leq k \leq n}(a+bn-bk)$$
Consider $\displaystyle \sum_{0 \leq k \leq n}(a+bk)$. Let $k_1 = n-k$. Hence, we get that $$\displaystyle \sum_{0 \leq k \leq n}(a+bk) = \sum_{0 \leq n-k_1 \leq n} (a+b(n-k_1))$$
Now $0 \leq n-k_1 \leq n$ is same as $0 \leq k_1 \leq n$. Hence, we get that $$\displaystyle \sum_{0 \leq k \leq n}(a+bk) = \sum_{0 \leq n-k_1 \leq n} (a+b(n-k_1)) = \sum_{0 \leq k_1 \leq n} (a+bn - bk_1)$$ Note that $k_1$ is just a dummy variable and hence you can as well replace it with $k$. Hence, $$\displaystyle \sum_{0 \leq k \leq n}(a+bk) = \sum_{0 \leq k \leq n} (a+bn - bk)$$
In general, the change of variable is the function $k_1 = p(k)$ in $$\sum_{k \in K}a_{k}=\sum_{k_1 \in K}a_{k_1}$$ Note that $p(k)$ is a permutation i.e. a bijection from $\{1,2,\ldots,n\}$ to $\{1,2,\ldots,n\}$. So $\sum_{k_1 \in K}a_{k_1}$ is just the same as $\sum_{k \in K}a_{k}$ just that the order in which you add the elements are different and thanks to commutativity, both remain the same. For instance, in the previous example, $k_1 = p(k) = n-k$, which is nothing but adding the elements in the reverse order.
Similarly, in the second summation, $$\sum_{1 \leq j < k \leq n}\frac{1}{k-j}=\sum_{1 \leq j < k+j \leq n}\frac{1}{k}=\sum_{1 \leq k \leq n}\sum_{1 \leq j \leq n-k}\frac{1}{k}$$
Let $k_1 = k-j$ i.e. $k = k_1 + j$. Hence, we get that $$\sum_{1 \leq j < k \leq n}\frac{1}{k-j}=\sum_{1 \leq j < k_1+j \leq n}\frac{1}{k_1+j-j}=\sum_{1 \leq j < k_1+j \leq n}\frac{1}{k_1}$$
Now we have $1 \leq j < k_1+j \leq n$. To understand this summation.
First let $k_1=1$. We then get that $1 \leq j < j+1 \leq n$. This gives us $1 \leq j \leq n-1$.
Now let $k_1 = 2$. We then get that $1 \leq j < j+2 \leq n$. This gives us $1 \leq j \leq n-2$.
Now let $k_1 = 3$. We then get that $1 \leq j < j+3 \leq n$. This gives us $1 \leq j \leq n-3$.
Hence, you see that for $1 \leq k_1 \leq n-1$, we find that $1 \leq j \leq n-k_1$. Hence, the summation becomes
$$\sum_{1 \leq j < k \leq n}\frac{1}{k-j}=\sum_{1 \leq j < k_1+j \leq n}\frac{1}{k_1+j-j}=\sum_{1 \leq j < k_1+j \leq n}\frac{1}{k_1} = \sum_{1 \leq k_1 \leq n-1} \sum_{1 \leq j \leq n-k_1}\frac{1}{k_1}$$
The reason we set $k_1$ a fixed value and look at the range of $j$ is because we want to sum over $j$ first as a function of $k_1$ and then sum over $k_1$. The reason we want to sum over $j$ first i.e. $\displaystyle \sum_{1 \leq j \leq n-k_1}\frac{1}{k_1}$, is that this sum is easy to evaluate is easier to evaluate. The inner sum i.e.$\displaystyle \sum_{1 \leq j \leq n-k_1}\frac{1}{k_1}$ is $\dfrac{n-k_1}{k_1}$.
You could also perform the summation in the reverse order. This will give you $$\sum_{j=1}^{n} \sum_{k_1 = 1}^{n-j} \frac1{k_1}$$ However, the inner sum is not easy to simplify as in the earlier case.
To conlude the post, interchanging summation/integrals is a very powerful technique in general and can be of real use to prove non-trivial results.