Find an exact sequence from a module $M$ to direct sum of localizations of $M$ in a finitely generated ring 
Suppose $f_1,f_2,\ldots, f_k\in R $ such that $R=(f_1,f_2,\ldots, f_k)$. Let $M$ be an $R$-module. Define $U(f)=(f,f^2,f^3,\ldots)$ and let
  $$M_i=U(f_i)^{-1}M$$
  $$M_{i,j}=U(f_if_j)^{-1}M$$
  where $U^{-1}M$ means the localization of $M$ with respect to $U$. Find a natural sequence
  $$0\rightarrow M\rightarrow \bigoplus_i M_i\rightarrow \bigoplus_{i,j}M_{i,j}$$
  and prove that this sequence is exact.

In my opinion, we have a natural map $\phi: M \rightarrow \bigoplus_i M_i$ by $m\mapsto (\frac{m}{f_1}, \ldots, \frac{m}{f_k})$. But I cannot find a "natural" map from $\bigoplus_i M_i$ to $\bigoplus_{i,j}M_{i,j}$. I want to construct a map so that the image of $\phi$ is the kernel of this map. But I don't see clearly how this is done. Any thought?
Also, here's a possibly related and potentially helpful question: Express the identity element as a linear combination of $N^{th}$ powers of generators in a ring.
 A: The map $M \to U(f)^{-1}M, m \mapsto \frac{m}{f}$ is NOT natural, since it depends on $f$, but for example $U(f)^{-1}M=U(f^2)^{-1}M$.
The natural map is $M \to U(f)^{-1}M, m \mapsto \frac{m}{1}$. Extending this to the direct sum situation, we have the natural map
$$M \to \bigoplus_{i} M_i, m \mapsto \left(\frac{m}{1}, \dotsc, \frac{m}{1}\right).$$
Let me prove injectivity of this map: $\left(\frac{m}{1}, \dotsc, \frac{m}{1}\right)=0$ means $\frac{m}{1} = 0 \in U(f_i)^{-1}M$ for any $i$, hence $f_i^{e_i} \in \operatorname{Ann}(m)$ for some $e_i$ by the definition of localization. This means $\operatorname{Ann}(m) \supset (f_1^{e_1}, \dotsc, f_k^{e_k})=R$, hence $m=0 \in M$. The fact $(f_1^{e_1}, \dotsc, f_k^{e_k})=R$ is proven in your linked related question.

To find a suitable map $\bigoplus_{i} M_i \to \bigoplus_{i,j} M_{i,j}$, note that for the exactness to hold, we need that $\left(\frac{m}{1}, \dotsc, \frac{m}{1}\right)$ is mapped to zero, so you should map $\left(x_1, \dotsc, x_k\right)$ to the direct sum of $x_i-x_j, 1 \leq i,j \leq k$.
Maybe you should prove the exactness of the sequence at this place by yourself.

Edit:
The exactness is not that easy. You can find the proof in any book an algebraic geometry, since this is basically the proof, that the structure sheaf on $\operatorname{Spec} A$ is a sheaf (Well at least in any book, that covers the modern language of schemes). It works like this:
Let $\left(\frac{m_1}{f_1^e}, \dotsc, \frac{m_k}{f_k^e}\right)$ be a tuple with $0=\frac{m_i}{f_i^e}-\frac{m_j}{f_j^e}=\frac{f_j^em_i-f_i^em_j}{(f_if_j)^e}$ for any $i,j$. Note that I can choose the same $e$ for all denominators, since I can always replace the numerator $m_i$ by some $f^{a}m_i$ to increase the exponent of the denominator.
By the definition of localization we get some $d > 0$, such that
$$0=(f_if_j)^d(f_j^em_i-f_i^em_j)$$
holds for any $i,j$ (Technically, we get some $d_{i,j}$ for any pair $i,j$, but we can pick the maximum, since there are only finitely many of that pairs).
With $t=d+e$, we can write this as
$$f_j^t(f_i^dm_i)=f_i^t(f_j^dm_j)$$
for any $i,j$. Now let $1=\sum_{i} a_if_i^t$ and set $m=\sum_i a_if_i^dm_i$. We compute for any fixed $j$:
$$f_j^{d+e}m=f_j^tm = \sum_i a_if_j^tf_i^dm_i = \sum_i a_if_i^tf_j^dm_j = f_j^dm_j,$$
or equivalently
$$f_j^d(m_j-f_j^em)=0,$$
which means $\frac{m}{1} = \frac{m_j}{f_j^e} \in U(f_j^{-1})M$. This holds for any $j$, hence our tuple is equal to $\left(\frac{m}{1}, \dotsc, \frac{m}{1}\right)$ and thus contained in the image of the first map.
