In the expression $\tag{1}\sum_{1\le k\le n+1} a_k$ the index set is the set of all $k$ such that $1\le k\le n+1$. This gives the index values $k=1,\quad k=2,\quad k=3,\quad \ldots\quad,\quad k=n+1.$ You then of course substitute these values into the summand $a_k$ and add: $ a_1+a_2+\cdots+a_{n+1}. $
For $\tag{2}\sum_{1\le k+1\le n+1} a_{k+1},$ the index set is the set of all $k$ such that $1\le k+1\le n+1$. This gives the index values $k=0,\quad k=1,\quad k=2,\quad \ldots\quad,\quad k=n. $ Substituting these values of $k$ into the summand $a_{k+1}$ gives the sum $ a_{0+1}+a_{1+1}+\cdots+a_{n+1}, $ which is the same as before.
In the sum $(2)$, you can think of $k+1$ as the index, since the summand $a_{k+1}$ plays along nicely with that interpretation. Doing this, $k+1$ takes values from $1$ to $n+1$ and in the summand $a_{k+1}$, you replace $k+1$ with these values.
You can take the index set to be whatever you like, as long as you adjust the summand appropriately so that the same sum results. For instance, you could write the sum as $\sum_{5\le k-3 \le n+5} a_{k-7}$ or as $\sum_{1\le k-7 \le n+1} a_{k-7}.$
But, writing as you suggest, $\sum_{1\le k+1 \le n+1} a_{k-1 } $ would give you $ a_{-1}+a_0+\cdots+a_{n -1.} $ You could, however, write the sum $(1)$ as $\sum_{2\le k \le n+2} a_{k-1} .$