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.
For a given $n$, this answer by r9m provides a tool to improve the upper bound for the left hand side by looking for the set of $\{x_k\}_{k=1}^n$ such that $c(n)=\max\limits_{k \in \{1,2,\cdots,n\}}\{c_k\}$ is minimized. It turned out to more convenient to consider a sequence $\{y_k\}_{k=2}^n$, where $y_{k}=x_k/x_{k-1}$ and $d(n)=2-c(n)$. Below I present the best values of $d(n)$ for small $n$ and the (componentwise almost optimal) sequences $\{y_k\}$ providing them which I found (the decimal fractions below are truncated, not rounded):
n=4 0.738307
2.195696 1.989904 2.218204
n=5 0.696237
2.071728 1.818972 1.821304 2.083424
n=6 0.660482
1.9875 1.7125 1.6625 1.7250 2.0000
n=7 0.632358
1.925 1.660 1.575 1.575 1.655 1.930
n=8 0.611866
1.8912 1.6148 1.5254 1.5012 1.5230 1.6106 1.8896
It suggests that there exists of a general pattern for $\{y_k\}$. Below are graphs for $\{y_k\color{red}{-1}\}$ for $n=20$ (which provides $d(20)\ge 0.4750702$) and $n=100$ (which provides $d(100)\ge 0.3156195$). I remark that the last value is more than twice bigger than $\frac{7\ln 2}{8\ln 100}=\frac{7}{8\log_2 100}=0.1317006\dots$, so I guess that analytic description of the pattern can lead to a proof of a strong inequality. As an aid I provide below the (rounded) sequence $\{y_k-1\}$ for $n=100$.
0.568813293
0.339731947
0.250658592
0.202100545
0.171168765
0.149589419
0.133606711
0.121254158
0.111401006
0.103346372
0.096632292
0.090945372
0.086065313
0.081830815
0.078121791
0.074847124
0.071934277
0.069329774
0.06698533
0.064867893
0.062946908
0.06119601
0.059597855
0.058134021
0.056788616
0.055549986
0.054408614
0.053355081
0.052381164
0.051477808
0.050642813
0.049869636
0.049150712
0.048487213
0.047870011
0.047300624
0.046773765
0.046287301
0.045839285
0.045426464
0.045050178
0.044705127
0.044392893
0.044110562
0.043858542
0.043635059
0.043439437
0.043271504
0.043130179
0.043015646
0.042927185
0.042864435
0.042829168
0.042819848
0.04283657
0.042881058
0.042952044
0.043050421
0.043179212
0.043337126
0.043524287
0.043744926
0.043999354
0.044286068
0.044612369
0.044977226
0.045381648
0.045832647
0.046329917
0.046878054
0.047482742
0.048146738
0.048877411
0.049679262
0.05056174
0.051533042
0.052601587
0.053782408
0.055087981
0.05653489
0.058146492
0.059945903
0.06196218
0.06423862
0.066818627
0.069766721
0.073161067
0.077108166
0.081748879
0.087280503
0.093985299
0.10227286
0.11278736
0.126570241
0.145453179
0.172979978
0.217060498
0.299864157
0.518985805
Best Answer
For the original first question where $l = k$, let $m=n=2$, $a_1=b_1=1$, and $a_2=b_2=2$; then
$$\sum_{k=1}^2a_k\sum_{k=1}^2b_k=\sum_{k=1}^2a_k(1+2)=1\cdot3+2\cdot3=9\;,$$
but $$\sum_{k=1}^2\sum_{k=1}^2a_kb_k=\sum_{k=1}^2(1^2+2^2)=5+5=10\;.$$
For the second question, imagine arranging the terms $a_ib_j$ in an $n\times m$ array:
$$\begin{array}{ccccc|c} a_1b_1&a_1b_2&a_1b_3&\dots&a_1b_m&\sum_{j=1}^ma_1b_j\\ a_2b_1&a_2b_2&a_2b_3&\dots&a_2b_m&\sum_{j=1}^ma_2b_j\\ a_3b_1&a_3b_2&a_3b_3&\dots&a_3b_m&\sum_{j=1}^ma_3b_j\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\ a_nb_1&a_nb_2&a_nb_3&\dots&a_nb_m&\sum_{j=1}^ma_nb_j\\ \hline \sum_{i=1}^na_ib_1&\sum_{i=1}^na_ib_2&\sum_{i=1}^na_ib_3&\dots&\sum_{i=1}^na_ib_m \end{array}$$
For each $j=1,\dots,m$, $\sum_{i=1}^na_ib_j$ is the sum of the entries in column $j$, and for each $i=1,\dots,n$, $\sum_{j=1}^ma_ib_j$ is the sum of the entries in row $i$. Thus,
$$\begin{align*} \sum_{j=1}^m\sum_{i=1}^na_ib_j&=\sum_{j=1}^m\text{sum of column }j\\ &=\sum_{i=1}^n\text{sum of row }i\\ &=\sum_{i=1}^n\sum_{j=1}^ma_ib_j\;. \end{align*}$$
For infinite double series the situation is a bit more complicated, since an infinite series need not converge. However, it is at least true that if either of
$$\sum_{j=1}^m\sum_{i=1}^n|a_ib_j|\quad\text{and}\quad\sum_{i=1}^n\sum_{j=1}^m|a_ib_j|$$
converges, then the series without the absolute values converge and are equal. This PDF has much more information on double sequences and series.