For (a), your resolution is not right, because $(\prod_{p\in S^1}i_p(\mathbb{Q}))/\mathbb{Z}$ is not $\prod_{p\in S^1}i_p(\mathbb{Q}/\mathbb{Z})$. Or in other words, the kernel of $\prod_{p\in S^1}i_p(\mathbb{Q})\rightarrow \prod_{p\in S^1}i_p(\mathbb{Q}/\mathbb{Z})$ is not $\mathbb{Z}$ (this is $\prod_{p\in S^1}i_p(\mathbb{Z})$).
For (b), you cannot mimick example 4.0.4 to compute the Cech cohomology. In fact, the higher Cech cohomology of this sheaf vanishes. Let $U,V$ your cover (which is good), and consider the Cech complex
$$ R(U)\times R(V)\rightarrow R(U\cap V)$$
given by $(f,g)\mapsto f_{|U\cap V}-g_{|U\cap V}$.
The kernel of this map are couple of functions $(f,g)$ that agree on the intersection, and hence patch together to form a unique function on $S^1$. So $\overset{\vee}{H^0}(\{U,V\},R)=R(S^1)$ as expected.
But I claim that the map is onto, so that $\overset{\vee}{H^1}(\{U,V\},R)=0$. Indeed, let $f\in R(U\cap V)$, also let $u,v$ be two functions on $S^1$ such that $u$ has (compact) support in $U$, $v$ has (compact) support in $V$ and $u+v=1$. In particular the function $u$ is zero in a neighborhood of $S^1\setminus V$ and $v$ is zero in a neighborhood of $S^1\setminus U$. The function $uf$ can then be extended to $U$ because it is zero in a neighborhood of $U\setminus U\cap V$. Similarily, the function $vf$ can be extended to $V$. And $(uf,-vf)\mapsto uf+vf=f$ so that the map is onto.
Here $u,v$ are called a partition of unity. A sheaf like $R$ with partitions of unity is called a fine sheaf. This argument (of a similar one with more than two open sets in the cover) shows that the higher Cech cohomology groups of a fine sheaf over a paracompact space vanish. On a paracompact space, Cech and derived functor cohomology agree so the cohomology of a fine sheaf is trivial.
The difference with the example 4.0.4 is that there is no partition of unity in a constant sheaf.
Copied from my answer on mathoverflow:
Suppose we can solve the problem for affine schemes, and choose an open affine $j: V \hookrightarrow Y$ and let $i: U \hookrightarrow X$ be its preimage. Suppose we have a morphism
$$ \alpha_V: \mathcal{O}_V^n \to (f|_U)_*\mathcal{O}_U$$
which is an isomorphism at the generic point $\eta$ of $Y$.
The problem is that $i_* \mathcal{O}$ is not in general coherent on $X$. But $\alpha_V$ chooses $n$ global sections $s_1,\dotsc,s_n \in \Gamma(X, i_* \mathcal{O}_U) = \Gamma(U, \mathcal{O}_U)$, which can be used to define a morphism $\alpha_X: \mathcal{O}_X^n \to i_* \mathcal{O}_U$. Let $\mathscr{G}$ be the image of this morphism. Then $\mathscr{G}$ is coherent, because for every open affine $\text{Spec }A = W \subset X$, $\mathscr{G}(W) \subset (i_*\mathcal{O}_U)(W)$ is the $A$-submodule generated by $s_1|_W,\dotsc,s_n|_W$.
This allows us to define the morphism
$$ \alpha_Y: \mathcal{O}_Y^n \to f_* \mathscr{G} \subset (f i)_* \mathcal{O}_U$$
by takting the same global sections $s_1,\dots,s_n \in \Gamma(Y, f_* \mathscr{G})$. At the generic point this yields
$$
\mathcal{O}^n_{Y, \eta} \xrightarrow{\alpha_{Y, \eta}} (f_*\mathscr{G})_\eta \hookrightarrow ((fi)_* \mathcal{O}_U)_\eta,
$$
and the composition is $\alpha_{U, \eta}$ which is an isomorphism. Hence $\alpha_{Y, \eta}$ is an isomorphism as well.
This is not a full answer, I just think that in the example $f: \mathbb{P}^1 \to \mathbb{P}^1, [x_0:x_1] \mapsto [x_0^2:x_1^2]$ the morphism $\alpha: \mathcal{O}_{\mathbb{P}^1}^2 \to f_*(\mathcal{O}_{\mathbb{P}^1}(1))$, which chooses the two generators of $\Gamma(\mathbb{P}^1, \mathcal{O}(1))$ does in fact work:
To differentiate the two instances of $\mathbb{P}^1$, let $x_0, x_1$ be the coordinates on the domain $X$ of $f$, and let $y_0, y_1$ be the coordinates on the image $Y = \mathbb{P}^1$. Then $\Gamma(X, \mathcal{O}(1)) = \langle x_0, x_1 \rangle_k$, and $\alpha: \mathcal{O}_{\mathbb{P}^1}^2 \to f_* \mathcal{O}_{\mathbb{P}^1}(1)$ is defined on global sections by sending the generator $e_i$ to $x_i$.
Choosing $U = D_+(y_1) \subset Y$, a principle open affine, we have $V = f^{-1}(U) = D_+(x_1)$. We want to study $\alpha$ restricted to $U$. Thus $U = \text{Spec } k\left[\frac{y_0}{y_1}\right]$, and $V = \text{Spec } k\left[\frac{x_0}{x_1}\right]$. The homomorphism of rings determining $f: V \to U$ is given by
$$ \frac{y_0}{y_1} \mapsto \left(\frac{x_0}{x_1}\right)^2.$$
The restriction $\langle x_0, x_1 \rangle_k = \Gamma(X, \mathcal{O}(1)) \to \Gamma(V, \mathcal{O}(1)|_V) = k\left[\frac{x_0}{x_1}\right]$ is given by mapping $x_0 \mapsto \frac{x_0}{x_1}$, and $x_1 \mapsto 1$.
If we consider $k\left[\frac{x_0}{x_1}\right]$ as a $k\left[\frac{y_0}{y_1}\right]$-module we see that it has two generators, namely $1$ and $\frac{x_0}{x_1}$. This is because with the $k\left[\frac{y_0}{y_1}\right]$-action we only hit the even degrees (or odd degrees, depending where you start).
Both generators are hit by $\alpha|_U$, which is therefor surjective on $U$-sections (I think it even is an isomorphism).
Best Answer
Fix $f\in K$ and $P\in\mathbb P^1$. You are looking for a $g\in K$ such that
$g\in\mathcal O_Q$ for all $Q\neq P$, and
$f-g\in\mathcal O_P$.
The first condition means that $g$ has poles only possibly at $P$. The second, that the difference $f-g$ does not have poles at $P$. In other words, $g$ is the singular part of $f$ at $P$. So to construct $f$ we may write the partial fraction decomposition of $f$ and drop all terms with a pole at a point different from $P$.