For finite $F$ the only difference between $\sum_{k\in F}a_k$ and $\sum_Fa_k$ is notational: the latter is an abbreviation for the former.
The theorem does indeed say that a non-negative series can be rearranged and broken up arbitrarily without any effect on its value. This is perhaps easiest to see in the special case in which $I=\Bbb Z^+$, say, so that
$$\sum_Ia_k=\sum_{k\in\Bbb Z^+}a_k=\sum_{k=1}^\infty a_k\;.$$
Call this sum $S$. Now we partition the positive integers into sets $I_j$ for $j\in\Bbb Z^+$: these sets are pairwise disjoint, and their union is all of $\Bbb Z^+$. The theorem says that if for each $j\in\Bbb Z^+$ we set
$$S_j=\sum_{I_j}a_k=\sum_{k\in I_j}a_k\;,$$
then
$$S=\sum_{j=1}^\infty S_j\;.$$
We might, for instance, let $I_1=\{2\}$, $I_2=\{1\}$, $I_3=\{4\}$, $I_4=\{3\}$, and so on, so that
$$I_j=\begin{cases}
\{a_{j+1}\},&\text{if }j\text{ is odd}\\
\{a_{j-1}\},&\text{if }j\text{ is even}\;;
\end{cases}$$
then
$$S_j=\begin{cases}
a_{j+1},&\text{if }j\text{ is odd}\\
a_{j-1},&\text{if }j\text{ is even}\;,
\end{cases}$$
and
$$\sum_{j=1}^\infty S_j=a_2+a_1+a_4+a_3+a_6+a_5+\ldots$$
is just the sum of the rearrangement in which we interchange $a_{2n-1}$ and $a_{2n}$ for each $n\in\Bbb Z^+$. Any rearrangement of the original series can be obtained in this way, and the theorem says that they all produce the same sum.
But it actually says even more than that, since the sets $I_j$ can contain more than one index. For instance, we can let
$$I_1=\{2n-1:n\in\Bbb Z^+\}=\{1,3,5,7,\ldots\}$$
and
$$I_2=\{2n:n\in\Bbb Z^+\}=\{2,4,6,8,\ldots,\}\;,$$
setting $I_j=\varnothing$ if $j>2$. Then
$$\begin{align*}
S_1&=\sum_{n=1}^\infty a_{2n-1}=a_1+a_3+a_5+\ldots\;,\\
S_2&=\sum_{n=1}^\infty a_{2n}=a_2+a_4+a_6+\ldots\;,
\end{align*}$$
and $S_j=0$ for $j>2$. Clearly we can ignore the $0$ terms, so the theorem says in this case that $S=S_1+S_2$, i.e., that
$$\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty a_{2n-1}+\sum_{n=1}^\infty a_{2n}\;:$$
we can sum the odd-numbered and the even-numbered terms separately, and the sum of those two subseries totals will be the same as the sum of the original series. Thus, the theorem covers not only rearrangements of the individual terms, but also breaking up the series into disjoint subseries and summing those subseries individually.
I can’t say exactly what is meant by the sum above is well-defined without having the entire relevant context. If it refers to $\sum_{k\in F}a_k$ for a finite $F$, for instance, it simply means that since this is a finite sum, we already know that the order in which it’s evaluated makes no difference, so we can safely specify the set of indices as a ‘lump’ — the set $F$ — instead of specifying the order in which the terms are to be added.
Best Answer
For your first question, the answer is yes, you can change indeed rewrite your outer sum by substituting $m-1$ for $j$ and changing the limits of that sum.
You can't, however, make the step indicated in your second question. Consider, for example, your middle summation. That sum is over all pairs $(m, j)$ satisfying by the predicate $$ P(m, j) = (1\le m\le 8) \land (1\le j\le 8) \land (j\ne m-1) $$ There will $64-8 = 56$ terms in this sum. The first step in your argument was correct: $$ \sum_{P(m, j)} 8 = 8\left(\sum_{P(m, j)} 1\right) $$ but your next step wasn't, since $$ \sum_{P(m,j)} 1 \ne \sum_1^8 1 $$ To see this, just observe that the sum on the left has 56 terms and the sum on the right has only 8.
Finally, a useful technique for problems like this is to ignore for the moment the excluded diagonal terms, sum all the terms, and then subtract the excluded ones. Using your chessboard idea, you noticed that the squares in the chessboard contain the numbers $1, \dots , 64$ so the sum of all the entries is $$ \frac{64\cdot65}{2} = 2080 $$ Now all we have to do is subtract the sum of the 7 diagonal elements in positions $$ (k-1, k) = (2, 1), (3, 2), (4, 3), (5, 4), (6, 5), (7, 6), (8, 7) $$ where the values are $9, 18, 27, 36, 45, 54, 63$. The sum of these is 252, so the answer to the problem is $2080-252=1828$.