Restricting split short exact sequence of quasi-coherent sheaves to their coherent sub-sheaves. Let $X$ be a projective variety, $Z$ a hypersurface section and $U \overset{def}= X \setminus Z$ its complement, an open affine subscheme of $X$. Let $i:U \hookrightarrow X$ be the corresponding open embedding. Given a coherent sheaf $M$ on $U$ we can consider the quasi-coherent sheaf $i_*M$. This seems to have a filtration of coherent sheaves in the following manner: Let's assume we are working on an affine chart and $W=\text{Spec}(R)$ is an affine in this chart and the equation of $Z$ corresponds to the element $f$ in $R$. The quasi-coherent sheaf $i_*M|_{W\cap U}$ on $W$ has this increasing filtration that $M_0=M|_W$, $M_1=\frac 1fM|_W$, $M_2=\frac 1{f^2}M|_W, \ldots$. This filtration seems to glue and give a filtration by coherent sheaves $(i_*M)_0\subset (i_*M)_1 \subset \ldots$ of $i_*M$. (Please correct me if I am wrong). Note that for each $j$, $i^*(i_*M)_j\cong i^*i_*M\cong M$.
Now let's consider a short exact sequence of vector bundles on $X$ like $0\rightarrow E_1 \rightarrow E_2 \rightarrow E_3 \rightarrow 0$. This leads to a short exact sequence of quasi-coherent sheaves $0\rightarrow i_*i^*E_1\rightarrow i_*i^*E_2\rightarrow i_*i^*E_3\rightarrow 0$ which restricts back to the original short exact sequence on $U$. Note that this exact sequence of quasi-coherent sheaves is split. This is because short exact sequence of vector bundles split on $U$. If the contents of the first paragraph is correct, my question is, does this split short exact sequence of quasi-coherent sheaves restrict to a split short exact sequence of coherent sheaves of the form $0\rightarrow (i_*i^*E_1)_j\rightarrow (i_*i^*E_2)_j\rightarrow (i_*i^*E_3)_j\rightarrow 0$ for each $j\geq 0$? (Here $j$ is the same indexing of the filtration defined in the first paragraph.)
 A: What you're doing when you compute $(i_*M)_j$ in the affine setting is that you have a module $M$ and you tensor up with $\frac 1f R$ over $R$, which is an $R$-submodule of $R_f \otimes_R M$. So in other words, $(i_*M)_j = \frac 1{f^j}R \otimes_R M$. I think your question can be converted into a module-theoretic question, i.e. is the $R$-module $\frac 1{f^j}R$ flat. I think your intuition for this statement comes from the fact that $\frac 1{f^j}R$ is an invertible $R$-module, since
$$
f^j R \otimes_R \frac 1{f^j} R \simeq R
$$
via the multiplication map, so that $\frac 1{f^j} R$ is an invertible $R$-module; this would be the module-theoretic equivalent of "twisting" when dealing with sheaves, in some sense I can't put my finger on since it's been a while. I'm not sure, but I think you may need the assumption that $f$ is a non-zero divisor so that the map $R \to R_f$ is injective and the computations in the above statement (i.e. the 'multiplication map') can happen within $R_f$.
Also because it's been a while, I don't feel super confident in my intuition that tells me that the above implies that $f^j R$ is faithfully flat for $j \in \mathbb Z$, but it seems to make sense when $X$ is integral, because then $R$ is integral, and saying that $f^jR$ is flat is basically saying that $f^jR_{\mathfrak p}$ is a free $R_{\mathfrak p}$-module of rank $1$ for any $\mathfrak p \in \mathrm{Spec}(R)$, and the isomorphism $R_{\mathfrak p} \to f^j R_{\mathfrak p}$ can clearly be given by multiplication by $f^j$ when both modules are seen as submodules of the quotient field $Q(R)$ (which is possible when $R$ is integral).
Hope that helps,
A: The answer to the second paragraph is negative. For example the zeroth filtration of the short exact sequence involving $i_*i^*E_k$'s just recovers the original short exact sequence. Every coherent sheaf $M$ such that $M|_U$ is isomorphic to a fixed coherent sheaf $E$, injects into the quasi-coherent sheaf $i_*E$. Applying the filtration corresponding to $M$, just reproduces the coherent sheaves $M, M(1), \ldots , M(n), \ldots$
