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.
Your analysis of the equivalence of (1) and (2) is correct. (2) and (3) are equivalent in the case that $X$ is noetherian, and it is the same idea, though covered slightly later in Hartshorne: III.3.6 says that every quasicoherent sheaf on a noetherian scheme can be embedded in a quasicoherent flasque sheaf, and flasque sheaves are acyclic for global sections, so one may compute cohomology using them. In the general case for $X$ not noetherian, the proof above does not apply and there ought to be counterexamples (see for instance this answer of Roland, though it does not contain an explicit counterexample).
The final paragraph contains a slight misconception: $X$ being an affine scheme only means that the global sections functor is exact on quasicoherent sheaves. We will show that we can have non-quasicoherent sheaves with higher cohomology on affine schemes via a dirty trick. Over an infinite field $k$, the underlying topological spaces of the schemes $\Bbb A^1_k$ and $\Bbb P^1_k$ are homeomorphic (both have one generic point and $|k|$ closed points, and are equipped with topologies where the closed sets are exactly the finite sets of closed points). Let $u:\Bbb A^1_k\to\Bbb P^1_k$ be such a homeomorphism. Then $H^i(\Bbb A^1_k,u^{-1}(\mathcal{F}))=H^i(\Bbb P^1_k,\mathcal{F})$ for $\mathcal{F}$ a sheaf of abelian groups, and so by picking $\mathcal{F}=\mathcal{O}(-2)$, for instance, we can find a sheaf of abelian groups with higher cohomology. The moral here is that we can't tell if a scheme is affine or not just from it's underlying topological space.
(2) should also be false but I'm unsure about a specific counterexample. I think you can use the example of the above paragraph by putting a funny $\mathcal{O}_X$-structure on the sheaf in question, but I'm having a brain-freeze about it right now.
Best Answer
Cohomology theories are those algebraic objects measuring the obstructions of some extending or glueing problems. One of the origins of Cech cohomology is the Mittag-Leffler problem.
Let $S$ be a Riemann surface, the Mittag-Leffler problem asks that given a discrete set of points $\left \{p_n \right \} \subset S$, whether we can find a meromorphic function $f: S \longrightarrow \mathbb{C}$ holomorphic outside $\left \{p_n \right \}$ and has predescribed principal part (the principal part of a Laurent series $\sum_{k=-N}^{\infty}a_k z^k$ is $\sum_{k = -N}^{-1}a_kz^k$) at each $p_n$?
This problem is locally trivial, so the thing is just how can we glue all the local solutions. We take an open covering $\bigcup_{i\in I}U_i = S$ such that each $U_i$ contains at most one $p_n$ and there is a local solution $f_i$ on $U_i$. Solving the problem globally is equivalent to finding $g_i \in \mathcal{O}(U_i)$ such that $f_{ij} = f_i - f_j = g_j - g_i$ (hence $f = f_i + g_i$ is a global solution).
This is precisely the definition of the first Cech cohomology $$\check{H}^1(\left \{U_i \right \},\mathcal{O}) = \frac{\left \{ \left \{f_{ij} \right \} \mid f_{ij}+f_{jk}+f_{ki} = 0 \right \}}{\left \{ \left \{f_{ij} \right \} \mid \exists g_i: f_{ij} = g_j - g_i \right \}}.$$ Note. Philosophically, higher cohomology groups are direct generalizations of $H^1$ and a way to compute $H^1$.