This question is a follow-up to this.
Let $A, B_1, B_2, \dots$ be countable sets such that $\bigcup_{i \in \mathbb{N}}B_i = A$ and the $B_i$'s are disjoint. Let $f : A \to \mathbb{R}$ take elements of $A$ to the reals. The claim is that $$\sum_{w \in A} f(w) = \sum_{i \in \mathbb{N}} \sum_{w \in B_i} f(w),$$ where the left-hand side converges if and only if the right-hand side converges. I am using this definition of $\sum_{w \in S} f(w)$:
We say that $\sum_{w \in S} f(w) = T$ (converges to $T$) if for each $\varepsilon > 0$, there exists finite subset $F \subseteq S$ such that for all finite $G$ satisfying $F \subseteq G \subseteq S$ we have $|\sum_{w \in G} f(w) - T| < \varepsilon$.
In the original question I only needed that left-hand side converges $\implies$ right-hand side converges. I wonder if the other direction can be proven too? Here is my proof of the first direction:
We first note that $\sum_{w \in B_i} f(w)$ converges for each $B_i$ since $B_i \subseteq A$.
Now let $\sum_{w \in A} f(w) = S$. Let $\varepsilon > 0$. So there exists finite $F \subseteq A$ such that for all finite $G$ satisfying $F \subseteq G \subseteq A$, we have $|\sum_{w \in G} f(w) - S| < \varepsilon/2$.
Let $J = \{i \in \mathbb{N} : F \cap B_i \neq \emptyset\}$ so $J \subseteq \mathbb{N}$ is finite. Now let $K$ be an arbitrary finite superset of $J$ such that $J \subseteq K \subseteq \mathbb{N}$.
For each $i \in K$, we can find finite $C_i \subseteq B_i$ such that $|\sum_{w \in C_i} f(w) - \sum_{w \in B_i} f(w)| < \frac{\varepsilon}{2|K|}$ and $B_i \cap F \subseteq C_i$. Thus,
$$\begin{eqnarray*} \left|\sum_{i \in K} \sum_{w \in B_i} f(w) - S\right| &\leq& \left|\sum_{i \in K} \sum_{w \in C_i} f(w) - \sum_{i \in K} \sum_{w \in B_i} f(w)\right| + \left|\sum_{i \in K} \sum_{w \in C_i} f(w) - S\right|\\ &<& |K|\frac{\varepsilon}{2|K|} + \varepsilon/2\\ &=& \varepsilon \end{eqnarray*}.$$
For the other direction, the main problem is finding the finite subset $F \subseteq A$. I am unable to find one that works. Is it provable?
The other direction is false. For a counterexample, let $B_i=\{i\}\times\mathbb N$, let $f((i,1))=1$ and $f((i,2))=-1$ and the remaining values any sufficiently convergent sequences, e.g. $f((i,j))=2^{-i-j}$.
Then the right-hand side converges, but there are infinitely many values of $1$, so you can't pick the required finite set for any $\epsilon$.