How does choice of sheaf affect what you’re measuring with cohomology? Can someone please explain, in intuitive terms, the difference between $H^1(X,\mathscr{O})$ and $H^1(X, \Omega)$ where $X$ is a Riemann surface, $\mathscr{O}$ is the sheaf of holomorphic functions on $X$, and $\Omega$ is the sheaf of holomorphic differential forms on $X$?
It might be illuminating to refer to the case of $X = \mathbb{P}^1$, in which case $H^1(X,\mathscr{O})$ is trivial and $H^1(X, \Omega)$ is a one-dimensional vector space generated (in some sense) by $dz/z \in \Omega(\mathbb{P}^1 \setminus \{0,\infty\})$.
First cohomology with respect to $\mathscr{O}$ measures the number of handles (the genus). The Riemann sphere has none of these. But what does the Riemann sphere have one of, according to $\Omega$?
 A: It's not straightforward to answer your question.  In general, when one writes the expression $H^i(X,\mathcal F)$, one tends to think ofhaving $X$ fixed, and $\mathcal F$ varying, and so one thinks of the cohomology as measuring properties of $\mathcal F$, rather than just of the space $X$.
In order to think of cohomology as measuring a property of $X$, rather than of $\mathcal F$, we have to choose an $\mathcal F$ that makes sense for all $X$.  One way is to take $\mathcal F$ to be a constant sheaf, like (the constant sheaf attached to) $\mathbb Z$ of $\mathbb C$.
Then we obtain $H^i(X,\mathbb Z)$ or $H^i(X,\mathbb C)$, which agrees with singular cohomology (for reasonable spaces $X$), and which has various well-known interpretations and heuristics in terms of the topological properties of $X$.
Another choice, when $X$ is a smooth variety, is to take $\mathcal F$ to be $\Omega_X^p$ for some $p \geq 0.$  (The case $p= 0$ gives $\mathcal O_X$, and the case $p = 1$ gives just $\Omega_X$ itself. )  
(Indeed, there aren't all that many other expressions for $\mathcal F$ that you can write down which are defined for an arbitrary smooth variety $X$.  One could also take the tangent bundle and its exterior powers, rather than the cotangent bundle, and one could take symmetric powers instead of exterior powers, or more general combinations of symmetric and exterior powers.  For example, $H^0$ of the powers of the canonical bundle, i.e. $\Omega_X^d$ where $d = \dim X$, give the so-called plurigenera of $X$.)
The dimension of $H^q(X,\Omega^p_X)$ is denoted $h^{p,q}$, and this collection of numbers are called the Hodge numbers of $X$.  They are the subject of what is called Hodge theory, and one key fact about them is that if $X$ is a smooth projective variety over $\mathbb C$, then
$$\sum_{p+q  = n} h^{p,q}  = \dim H^n(X,\mathbb C).$$
E.g. for a smooth projective curve $X$, we find that 
$$h^{1,1} = \dim H^1(X,\Omega^1_X)  = \dim H^2(X,\mathbb C) = 1,$$
the last equality following because as a manifold $X$ is compact and oriented of dimension two, and so its $H^2$ is one-dimensional, spanned by the fundamental class.  So the answer to your question "what does the Riemman sphere [or any curve] have one of" is "it has one fundamental class".
Note also that for a genus $g$ curve, $h^{1,0} = h^{0,1} = g,$ and these
add up to give $2g$, which is the dimension of $H^1(X,\mathbb C)$.  (In general, Hodge symmetry says that $h^{p,q} = h^{q,p}$.)
Summary/conclusion: perhaps the most interesting and suggestive answer to your question comes from Hodge theory, and so maybe this is a direction you should explore.
