Let's start with something more general:
Assume, that for some finite set $I$, you have a sum
$$
\sum_{l \in I} s_i
$$
Now, the summands correspond to the elements in $I$. If, for some set $J$, we have a bijection
$$
I \overset{\sigma}{\longrightarrow} \bigcup_{j \in J}I_j =: Y
$$
where $I_j \cap I_i = \emptyset$ for $i\neq j$, then
$$
\sum_{i\in I} s_i = \sum_{y \in Y} s_{\sigma^{-1}(y)} =
\sum_{y \in \bigcup_{j \in J}I_j} s_{\sigma^{-1}(y)} =
\sum_{j \in J}\sum_{y \in I_j} s_{\sigma^{-1}(y)}
$$
The first equation holds because $\sigma$ is a permutation and addition is commutative. The second equation is trivial and the third equation uses that the union is disjoint, as mentioned above.
So, any bijection $\sigma$ into a disjoint union as mentioned above will give you a representation as a double sum. You could of course proceed inductively, to get triple, quadruple sums et cetera.
The double sum can hence be understood as a specific grouping of summands/a partition of the index set.
Concerning your special case:
Defining $I:=\left\{\left(j,k\right);\,1\leq j < k \leq N\right\}$, you can write
$$
I = \bigcup_{1 \leq j < N} \underbrace{\left\{(j,k);\, j < k \leq N\right\}}_{:= I_j}
$$
and the union is disjoint. Taking $\sigma$ as the identity, and using the above, we get
$$
\begin{eqnarray}
& 2 \sum_{(j,k) \in I} P(F_j F_k) =
2 \sum_{(j,k) \in \bigcup_{1 \leq j < N} I_j} P(F_j F_k) =
2 \sum_{1 \leq j < N} \sum_{(j,k) \in I_j} P(F_j F_k) = \\
& 2 \sum_{1 \leq j < N}\; \sum_{(j,k) \in \left\{(j,m);\, j < m \leq N \right\}} P(F_j F_k) = 2 \sum_{1 \leq j < N} \sum_{j < k \leq N} P(F_j F_k) \\
\end{eqnarray}
$$
where in the last equation, we used that the sets
$
\left\{(j,m);\, j < m \leq N\right\}
$
and
$
\left\{m;\, j < m \leq N\right\}
$
are bijective/have the same number of elements. There is some subtle formal detail involved in this last equation:
We are basically applying what we used before - $\sum_{i\in I} s_i = \sum_{y \in Y} s_{\sigma_j^{-1}(y)}$ - to the inner sum, where for each $j$, $\sigma_j: \left\{(j,m);\, j < m \leq N\right\} \longrightarrow \left\{m;\, j < m \leq N\right\},\; (j,m) \mapsto m$ is the projection to the second component and $\sigma_j^{-1}$ is the function that maps $m \mapsto (j,m)$.
Explicitly,
$$
\begin{array}
& 2 \sum_{1 \leq j < N}\; \sum_{(j,k) \in \left\{(j,m);\, j < m \leq N \right\}} P(F_j F_k) =
2 \sum_{1 \leq j < N}\; \sum_{(j,k) \in \left\{(j,m);\, j < m \leq N \right\}} P(F_j F_{\sigma_j(j,k)}) = \\
2 \sum_{1 \leq j < N} \sum_{j < k \leq N} P(F_j F_{\sigma_j(\sigma_j^{-1}(k))}) =
2 \sum_{1 \leq j < N} \sum_{j < k \leq N} P(F_j F_k) \\
\end{array}
$$