If you're having trouble with the shift in summation you can put the "missing" step back in, which should help a little. Starting with your first example, we have:
$$
A(n):= \sum_{k=n}^{2n-1} \frac{1}{1+k} = \sum_{j=??}^{j=??} \frac{1}{j}
$$
and we need to determine what the new limits in the sum are. So, when $k=n$ we have $1/(n+1)$ as the summand, which now needs to be $1/j$. So $j$ must be $n+1$. Likewise, when $k=2n-1$ we have $1/(1+2n-1)=1/(2n)$ in the summand, so $j=2n$. So we transform $A(n)$ into
$$ A(n) = \sum_{j=n+1}^{2n} \frac{1}{j} $$
and then finally we replace $j$ with $k$ to get your version. Reusing the $k$ is a standard mathematical technique, partly to avoid using too much notation and partly to keep the theme of certain elements in the sum present to convey the main idea.
For the second part of your question, the sum is being rewritten with certain of the summation terms written out explicitly. Consider:
$$\sum_{i=1}^{n+1} i = \left(\sum_{i=2}^n i\right) + 1 + (n+1)$$
Here, we've removed the terms $i=1$ and $i=(n+1)$ from the sum and written them explicitly after the sum. That's exactly what's happening for you above.
So:
$$A(n+1) = \sum_{k=n+1}^{2n} \frac{1}{1+k} = \sum_{k=n}^{2n-1} \frac{1}{k+1} + \frac{1}{2n} - \frac{1}{1+n} $$
because we have to add on the "missing" last term $1/(k+1)$ because the index $k$ now only goes as far as $n$ instead of $n+1$, and we subtract the "extra" term $1/n$ because the lower index now goes down to $n$ instead of stopping at $n+1$.
It might help if you set $n$ to be a small value, say $2$ or $3$ and write the sums out in full to see all the terms appear.
