Map between free sheaf modules not arising from a "matrix" My question is about this example in  the Stacks project.
For convenience and completeness, here is the relevant part.

Let $X$ be countably many copies $L_1, L_2, L_3, \ldots$
  of the real line all glued together at $0$; a fundamental
  system of neighbourhoods of $0$ being the collection
  $\{U_n\}_{n \in \mathbf{N}}$, with $U_n \cap L_i = (-1/n, 1/n)$.
  Let $\mathcal{O}_X$ be the sheaf of continuous real valued functions.
  Let $f : \mathbf{R} \to \mathbf{R}$ be a continuous function
  which is identically zero on $(-1, 1)$ and identically $1$
  on $(-\infty, -2) \cup (2, \infty)$. Denote $f_n$ the continuous
  function on $X$ which is equal to $x \mapsto f(nx)$ on each
  $L_j = \mathbf{R}$. Let $1_{L_j}$ be the characteristic function
  of $L_j$. We consider the map
  $$
\bigoplus\nolimits_{j \in \mathbf{N}}
\mathcal{O}_X
\longrightarrow
\bigoplus\nolimits_{j, i \in \mathbf{N}}
\mathcal{O}_X, \quad
e_j \longmapsto \sum\nolimits_{i \in \mathbf{N}} f_j 1_{L_i} e_{ij}
$$
  with obvious notation. This makes sense because this sum is locally
  finite as $f_j$ is zero in a neighbourhood of $0$.

For a fixed $j$, I see why the sum $\sum_{i\in \mathbb{N}} f_j 1_{L_i} e_{ij}$ is locally finite.
Every point that is not the origin admits a neighbourhood $U$ that lies completely in $L_k$ for some $k$. Then the sum reduces to the term  $f_k e_{kj}$. 
And $U_j$ provides a neighbourhood of the origin where the sum is finite.
The sum is not only finite, but the zero function on this nhood.
This does not use the fact that $f_j$ depends on $e_j$. It would also work if I use $f_3$ in every sum. 
What is the error in this reasoning? Or is it really redundant. 

I think I have found your misunderstanding:
The sum is not finite on $U_n$. Observe that $e_j$ maps to an infinite collection of functions $(g_{i,j})$ on $X$, where $g_{i,j}$ takes the value $f_j$ on $L_i$ and is zero on all the other lines. Since $U_n$ is the union of the copies of $(-1/n,1/n)$ in every line $L_i$, the image of $e_m$ for any $m>2n$ will be an infinite collection of non-zero functions on $X$, each one is identically $1$ on the union of intervals $(-1/n,-2/m)\cup(2/m,1/n)$ contained in one of the lines. It is exactly the fact that the $f_j$ "converge" to a function that is $0$ at the origin and $1$ everywhere else that show no matter how small a neighbourhood of the origin you take, some $e_m$ will always map to an infinite sum of terms. I hope this helps.
Also, where you have written "The sum reduces to the term $f_ke_{ik}$", I think you should instead have $f_ke_{kj}$.
