I'm probably overlooking something very trivial. Suppose that $A$ is some ring, $M \hookrightarrow \bigoplus_{j \in J} A$ a left $A$-submodule and $J$ is well-ordered. Let $F_j := \bigoplus_{i;i \leq j} A$ and $F_j' := \bigoplus_{i;i < j} A$, then by projecting on the last coordinate $M \cap F_j/M \cap F_j'\cong I_j$, some left ideal.
Now, there's an epi $\bigoplus_{j \in J} M \cap F_j \twoheadrightarrow M$ given by $f((m_j)_j) = \sum_{j} m_j$. Clearly, if $f((m_j)_j) = 0$, then $p_{j_n} (f((m_j)_j)) = 0$ ($p_{j_n}$ is the projection onto the $j_n$-th coordinate), where $j_n$ is the highest $j$ such that $m_j$ is non-zero. By induction, the kernel is $\bigoplus_{j \in J} M \cap F_j'$. However cokernels commutes with coproducts so $$\bigoplus_{j \in J} M \cap F_j/ \bigoplus_{j \in J} M \cap F_j'\cong \bigoplus_{j \in J} M \cap F_j/M\cap F_j'\cong \bigoplus_{j \in J} I_j.$$
This should be true for hereditary rings, but not for an arbitrary ring (see, for instance, Every Submodule of a Free Module is Isomorphic to a Direct Sum of Ideals). There's, then, a mistake above. I'm ashamed to admit it, but I can't find it.
Thanks in advance.
The problem seems to be in the displayed equation, where you have changed the embeddings of the submodules.
Here is an explicit example. In the linked answer one considers the ring $A=k[x,y]/(x^2,y^2)$, where $k$ is a field, and the submodule $M$ of $A^2$ generated by $(x,y)$. Thus $M$ has $k$-basis $(x,y),(xy,0),(0,xy)$.
Now $M\cap F_1=k(xy,0)=:N$ and $F_2=A^2$. Your epimorphism is then $f\colon N\oplus M\to M$, with kernel $K$ isomorphic to $N$ via the embedding $(1,-1)\colon N\to N\oplus M$.
We see that $(N\oplus M)/K\cong M$, obviously, but this is not isomorphic to $N\oplus(M/N)$, which is what you are claiming in the displayed equation.