Morphism between quotient sheaf ($K_{X}/\mathcal{O}_{X}$) and $\bigoplus_{x\in\mathcal{X}}i_{x,*}(K(X)/\mathcal{O}_{X,x})$ Consider an integral scheme $X$ of finite type over $k$ of dimension $1$. Then $X$ consists of one generic point $\eta$ and the rest of the points are closed points. We call the set of closed points $\mathcal{X}$. Let $K_{X}$ be the constant sheaf with value $K(X)(=\mathcal{O}_{X,\eta})$. I want to show that the map 
$$K(X)\rightarrow \bigoplus_{x\in\mathcal{X}}K(X)/\mathcal{O}_{X,x}, \qquad [U,f]\mapsto ([U,f] + \mathcal{O}_{X,x})_{x\in\mathcal{X}}$$ is a well-defined map of $k$-vectorspaces. 
Background information: By showing that this is a well-defined map of $k$-vectorspaces one can actually show that the quotient sheaf $K_{X}/\mathcal{O}_{X}$ is isomorphic to the direct sum sheaf $\bigoplus_{x\in\mathcal{X}}i_{x,*}(K(X)/\mathcal{O}_{X,x})$ induced by the inclusion $i_{x}:\{x\}\rightarrow X$. It is clear that this particular direct sum sheaf is flasque. We can thus find a canonical flasque resolution of $\mathcal{O}_{X}$ to compute the cohomology groups of $\mathcal{O}_{X}$.
Problems: 
1. To show that this map is well-defined we need for every $[U,f]\in K(X)$ that for only finitely many $x\in\mathcal{X}$, $[U,f]\neq 0$. But I don't see why this has to be the case. 
2 I struggle with the scalar multiplication inside $K(X)$. I know that $\mathcal{O}_{X,\eta}\cong \mathcal{O}_{\operatorname{Spec}(R),\eta} = \operatorname{Frac}(R)$ for some finitely generated $k$-algebra $R$ by the fact that $X$ is of finite type over $k$. Thus $K(X)$ indeed is a field and has a $k$-structure on it. But I don't understand the structure well enough to show (formally) that the map indeed preserves the scalar multiplication. 
Any help would be appreciated! 
 A: First, we reduce to the affine case: the complement of an affine open $U\subset X$ is a finite set of closed points, so it's enough to prove that the image of an element in $K(X)$ is nonzero in finitely many $K(X)/\mathcal{O}_{X,x}$ as $x$ ranges over the closed points in $U$. But this is exactly the statement for affine $X$.
Now suppose $X$ is affine with coordinate algebra $k[X]$. For an arbitrary element $f\in k(X)$, pick a representation $f=g/h$ with $g,h\in k[X]$ and $h\neq 0$. For a closed point $x\in X$, $f$ belongs to $\mathcal{O}_{X,x}$ if $h$ does not vanish at $x$. But $h$, being nonzero, vanishes on a finite set of closed points. So $f\notin \mathcal{O}_{X,x}$ for only finitely many $x\in X$, and thus we really do have a map in to the direct sum instead of the direct product.
I'm not totally sure what you're getting at with your second question. If one has $A$ an $R$-algebra, then any localization of $A$ is an $R$-algebra too: if we have an element $\frac{f}{s}\in S^{-1}A$, then $r\cdot \frac{f}{s}=\frac{rf}{s}$, and in this way the image of $A$ in $S^{-1}A$ by the map $a\mapsto \frac{a}{1}$ is naturally a sub-$R$-module.  In our case, we have that $k(X)$ is a localization of $\mathcal{O}_{X,x}$ and the natural map $\mathcal{O}_{X,x}\to k(X)$ is an injection, so our quotient $k(X)/\mathcal{O}_{X,x}$ is a quotient of $k$-modules and the quotient map $k(X)\to k(X)/\mathcal{O}_{X,x}$ is a map of $k$-modules. Does this clear things up?
A: I think (please feel free to correct me) $K(X)$ is basically the field of fractions and the direct sum is the residue fields of "values" your functions take at each point. So basically you have the evaluation map which should be a well-defined map of $k$-vector spaces.
