New answer:
I'm going to explain in what sense the answer is yes, and why the "counterexample" I gave in my last answer was contrived.
First, recall that the cohomology $H^i(X, F) = R^i\Gamma(X, F)$ is only defined up to natural isomorphism, and depends on the choice of injective resolution of $F$. So, let's denote by $H^i(F,I)$ the cohomology with respect to an injective resolution $I$ of $F$.
If we pick a different resolution $J$ there is a natural map $H^i(F,I) \cong H^i(F,J)$ which doesn't depend on all the choices you make to construct it. So, if $I$ happens to be a resolution by $\mathcal O$-modules, this makes $H^i(F,I)$ an $\mathcal O$-module, and even if $J$ is not a resolution by $\mathcal O$-modules the natural isomorphism gives its cohomology an $\mathcal O$-module structure.
Since the map from the Cech sheaf to an injective resolution also gives natural maps $\check H^i(X,F) \to H^i(F,I)$ and $\check H^i(X,F) \to H^i(F,J)$, these will respect both the $\mathcal O$-action on $I$ and on $J$.
So in this sense the answer is yes, the map is always an isomorphism of $\mathcal O$-modules.
The example I gave originally is actually pretty contrived. It arose from this observation: If you take a complex of $\mathcal O$-modules $J$ which is an injective resolution of $F$ as abelian groups, but for which the augmentation $\varepsilon : F \to J$ is not $\mathcal O$-linear, then the map $\check H^i(X,F) \to H^i(F,J)$ will not be $\mathcal O$-linear, nor will the natural isomorphism $H^i(F,I) \cong H^i(F,J)$ with respect to be $\mathcal O$-actions.
Original answer:
How do we prove that sheaf cohomology and cech cohomology are the same anyway? For reference see a nice note on this here: pub.math.leidenuniv.nl/~edixhovensj/teaching/2011-2012/AAG/lecture_14.pdf or see Hartshorne.
Basically you can build a resolution of $\mathcal F$ by making a sheaf-version of the Cech complex (whose global sections is the usual Cech complex and which is a sheaf of $\mathcal O_X$-modules). After you check this is a resolution, it is a fact that any resolution will map to an injective resolution. This involves making choices, but since the choices are unique up to homotopy you get the well-defined map from Cech cohomology to sheaf cohomology. But what if you switch the category from $\mathcal O_X$-modules to sheaves of abelian groups - are the choices still unique up to homotopies taken from the other category?
The answer is no. Consider the case $X$ is a point. Let $\mathcal F = \mathcal O_X = \underline{k}$ be the constant sheaf. If we compute in the category of sheaves of abelian groups then we have the freedom to replace $\mathcal F$ with an injective resolution. One injective resolution is $\phi: \underline{k} \to \underline{k}$ taken to be any isomorphism chosen specifically not to be $k$-linear but only additive. Then in this case the map between sheaf cohomology and cech cohomology is not $\mathcal O$-linear by design because it coincides with $\phi$ itself.
So, if you want an $\mathcal O$ linear map you can get one for free, and I think it is pretty standard to assume this map is $\Gamma(\mathcal O,X)$-linear.
Best Answer
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.