Why is this de Rham 1-cochain a cocycle on an elliptic curve?

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I am trying to understand a proof about the de Rham cohomology of an elliptic curve in Kedlaya's notes, p7 Example 1.5. The motivation for this question has overlap with my previous question

Let $X = \mathrm{Proj}K[X,Y,W]/(Y^2W-X^3-aXW^2-bW^3)$ be an elliptic curve over a field of characteristic zero, $U = X\setminus\infty$ the complement of infinity, and $V$ the complement of the three nontrivial 2-torsion points.

We compute the de Rham cohomology $H^i_\mathrm{dR}(X)$ by a $\check{C}$ech resolution as follows: $$D^0 = \Gamma(U,\mathcal{O}_U) \oplus \Gamma(V,\mathcal{O}_V)$$ $$D^1 = \Gamma(U,\Omega_U)\oplus\Gamma(V,\Omega_V)\oplus\Gamma(U\cap V, \mathcal{O}_{U\cap V})$$ $$D^2 = \Gamma(U\cap V, \Omega_{U\cap V})$$

To compute $H^1_\mathrm{dR}(X)$, note that the 1-cochains are triples $(\omega_U,\omega_V,f) \in D^1$, and the 1-cocycles are those such that $df = \omega_V-\omega_U$, and the 1-coboundaries are of the form $(df,dg,g-f)$.

We aim to show that the map $(\omega_U,\omega_V,f) \mapsto \omega_U$ induces a bijection $$H^1_\mathrm{dR}(X) \to H^1_\mathrm{dR}(U)$$

My questions regard the argument for injectivity. To that end, let $(\omega_U,\omega_V,f) \in D^1$ be a cocycle, and suppose it maps to 0, i.e. that $\omega_U$ is a 1-coboundary on $U$. Kedlaya then claims that $(0, \omega_V,f)$ is also a 1-cocycle.

However, for $\omega_U$ to be a 1-coboundary on $U$ means that $\omega_U = dg_U$ for some $g_U \in \Gamma(U,\mathcal{O}_X)$. Then $(\omega_U, 0, -g_U)$ is a 1-coboundary on $X$. So then I would expect we could conclude $(0,\omega_V,f+g_U)$ is a 1-cocycle.

Question: Why can we conclude that $(0,\omega_V,f)$ is a 1-cocycle?

Then he writes that, since $(0, \omega_V, f)$ is a 1-cocycle, we have $df = \mathrm{Res}_{V,U\cap V}(\omega_V)$, which means $f$ cannot have a pole inside $V$ or otherwise $df$ would have a double pole there. Therefore $\omega_V$ is a 1-coboundary of $V$, and hence $(0,0,f)$ is a 1-cocycle on $X$.

Question: Saying that $f$ has no pole in $V$ is a way of saying that $f \in \mathcal{O}(U\cap V)$ lifts to an element of $\mathcal{O}(V)$, correct? Would this also hold if we replaced $f$ by $f + g_U$? This argument about poles feels a bit hand-wavy to me. Is there a more sheaf-theoretic way of saying it?

Then he concludes that $\omega_V$ is a 1-coboundary on $V$, so $(0,0,f)$ is a 1-cocycle on $X$. I have the same issue here as when he concluded that $(0,\omega_U,f)$ is a 1-cocycle.

Lastly he says that $(0,0,f)$ being a 1-cocycle is only possible if $f$ is constant. This makes sense to me, but is there a more rigorous way of explaining it (e.g. sheafly)? (question)

I appreciate any help.