A description via cocycle usually is not convenient to work with. From my point of view, a description as an extension is much more useful. But if you want something else, I would advice the following. Since your curve $C$ is given as a double covering of $P^1$, that is as a relative spectrum of the sheaf of algebras $A = O \oplus O(-2)$ on $P^1$ (with $O$ summand being generated by the unit and with multiplication $O(-2)\otimes O(-2) \to O$ given by the ramification divisor $D$), the category of coherent sheaves on $C$ is identified with the category of sheaves on $P^1$ with a structure of a module over this algebra.
Spelling this out, a sheaf on $C$ is a sheaf $F$ on $P^1$ with a morphism $\phi:F(-2) \to F$ such that the composition $\phi\circ\phi(-2):F(-4) \to F(-2) \to F$ coincides with the morphism given by $D$. The structure sheaf $O_C$ corresponds to $A$. Hence the extension of $O_C$ with $O_C$ corresponds to an extension of $A$ with $A$, that is we have an exact sequence
$$
0 \to O \oplus O(-2) \to F \to O \oplus O(-2) \to 0.
$$
Since $Ext^1_A(A,A) = Ext^1_O(O,A) = H^1(A) = H^1(O \oplus O(-2))$, we see that the nontrivial extension corresponds to the extension of $O$ in the right term by $O(-2)$ in the left term. Thus $F = O \oplus O(-1)^2 \oplus O(-2)$ and the above sequence can be rewritten as
$$
0 \to O \oplus O(-2) \to O \oplus O(-1) \oplus O(-1) \oplus O(-2) \to O \oplus O(-2) \to 0
$$
The maps are given by matrices $\left[\begin{smallmatrix}1 & 0\cr 0 & x\cr 0 & y\cr 0 & 0 \end{smallmatrix}\right]$ and $\left[\begin{smallmatrix} 0 & y & -x & 0 \cr 0 & 0 & 0 & 1 \end{smallmatrix}\right]$. The map $\phi:F(-2) \to F$ providing $F$ with a structure of an $A$-module is given by the matrix $\left[\begin{smallmatrix} 0 & g & h & -f \cr x & xy & -x^2 & h \cr y & y^2 & -xy & -g \cr 0 & y & -x & 0 \end{smallmatrix}\right]$. Here $f$ is the polynomial in $x$ and $y$ of degree $4$ such that $div(f) = D$, and $g,h$ are polynomials such that $f = gx +hy$.
As Piotr remarks, these kind of questions quickly lead to studying reflexive sheaves. I would add that one also better get acquainted with Serre's condition $S_2$. For more on this see
this and this MO answers.
Perhaps the first remark is that besides the singularities of the surface $X$ you also have to take into account the singularities of the sheaf you are considering.
You are asking about normal surfaces. Normal implies $S_2$ and pretty much everything that follows is OK for $S_2$ in arbitrary dimensions.
A coherent sheaf on an open set can always be extended as a coherent sheaf on the ambient space. Furthermore if $X$ is $S_2$ and $j:U\hookrightarrow X$ is an open set such that $\mathrm{codim}_X(X\setminus U)\geq 2$, then a coherent sheaf $\mathscr F$ on $X$ such that $\mathscr F|_U$ is locally free is reflexive if and only if $$\mathscr F \simeq j_*(\mathscr F|_U).$$
Notice that this means that if $X$ is $S_2$, then a locally free sheaf $\mathscr E$ on $U$ can be extended as a locally free sheaf to $X$ if and only if $j_*\mathscr E$ is locally free. (The point is, that since $X$ is $S_2$, $j_* \mathscr E$ is reflexive and if there is a locally free sheaf extending $\mathscr E$, then that is also a reflexive sheaf which agrees with $j_*\mathscr E$ on an open set with at least codimension $2$ complement, so they have to agree).
This also tells you how to produce locally free sheaves that cannot be extended as a locally free sheaf: Take any reflexive sheaf that is not locally free, then take the open set where it is locally free. By the above, this sheaf on that open set is a locally free sheaf that does not have an extension as a locally free sheaf on the ambient space. Piotr's example is probably the simplest such sheaf.
So now the question is: When is a reflexive sheaf locally free?
In some contexts one defines the singularity set of a coherent sheaf $\mathscr F$ as the locus where it is not locally free. Then there are various results that say that the singularity set of certain sheaves have to be at least such and such codimension. Here is a short list of those:
Sample statement: If $\mathscr F$ is bluh, then the singularity set of $\mathscr F$ has codimension at least boo.
Actual statement If $X$ is smooth, we can substitute the following into the above sentence:
- If bluh = coherent, then boo = $1$.
- If bluh = torsion-free, then boo = $2$.
- If bluh = reflexive, then boo = $3$.
You may recognize theorems that you know as simple consequences of these:
Thm 1 For any coherent sheaf there exists a non-empty open set on which it is (locally) free.
Remark You may also note that this does not require $X$ smooth, but that's essentially because the conclusion is invariant under first restricting to the open set where $X$ is smooth.
Thm 2 A torsion-free sheaf on a smooth curve is locally free.
Remark We know that this fails if the variety is either not smooth or has dimension at least $2$.
Thm 3 A reflexive sheaf on a smooth surface is locally free.
Remark This is the reason why the statement you mention at the start is true. If $X$ is a smooth surface, then a sheaf that's locally free on the complement of finitely many points pushes forward to a reflexive sheaf which by this theorem has to be locally free.
Piotr's example or any Weil divisor that's not Cartier shows that this fails if the surface is not factorial. (Meaning that every local ring is a UFD. This is a weaker condition than being smooth.) It is also true that a reflexive sheaf of rank $1$ on a smooth variety is always locally free, so you need to go to higher ranks to get a counter-example.
To complete the picture here is an example of a reflexive sheaf on a smooth $3$-fold that is not locally free.
Example Consider the Euler sequence of $\mathbb P^3$:
$$
0\to \mathscr O_{\mathbb P^3}\to \bigoplus_{i=0}^3 \mathscr O_{\mathbb P^3} (1) \to
T_{\mathbb P^3} \to 0
$$
and consider (one of) the induced maps
$$
\alpha: \mathscr O_{\mathbb P^3} (1) \to T_{\mathbb P^3}.
$$
This is clearly injective and let the cokernel of $\alpha$ be $\mathscr F$.
The original (Euler) sequence shows that $\alpha$ cannot be a vector bundle embedding and hence $\mathscr F$ cannot be locally free. I leave it for you to prove that it is reflexive. (This is not absolutely trivial, but it is a good exercise).
So, to answer your question, the statement you cite is not true for singular varieties, at least not in the way it is stated. You can cook up some versions that work by adding additional assumptions. In any case, these are likely the notions that will help you do that.
Finally, here are some references:
- On reflexive sheaves I would recommend Hartshorne's series of papers Stable reflexive sheaves I-II-III.
- You can read about the singularity set and other interesting stuff in Chapter 2 of Okonek-Schneider-Spindler's Vector bundles on complex projective spaces
Best Answer
The kernel $E'$ of $u_s \colon E \to F$ is a torsion free sheaf over the curve $C$, hence it is necessarily a vector bundle (torsion free sheaves over curves are locally free).
Since the generic rank of $E'$ is $2$, it follows that $E'$ is a rank $2$ vector bundle.
The point you are missing is that the inclusion $E' \to E$ is not an inclusion of vector bundles, but only an inclusion of sheaves.
In fact, we can have a similar situation in rank one, for instance
$0 \to \mathcal{O}_C(-s) \to \mathcal{O}_C \to \mathbb{C}_s \to 0.$
Also in this case, the kernel of $v_s \colon \mathcal{O}_C \to \mathbb{C}_s$ is locally free.