Euler characteristic of a curve Is it a complete coincidence that the Euler characteristic of a curve is exactly twice the Euler characteristic of its structure sheaf?
 A: The answers that this follows from Hodge theory and HRR are of course spot on.  If what is meant is rather why those theorems are true, here is a short proof at least of HRR for curves:
If two smooth plane curves have the same degree, hence the same topology, they are linearly equivalent, so by the exact sheaf sequence (of a divisor on a surface) also have the same chi(O).  So chi(O) is a topological invariant for smooth plane curves.  But why does it equal 1-genus?
In degree one, both (1-genus) and chi(O) are equal to 1.  Moreover when you add a general line L to raise the degree of a curve C from n to n+1, the topological genus of the (smoothed) curve goes up by n-1, so (1-genus) goes down by n-1.  By the exact sequence relating the structure sheaf of the sum C+L of these divisors, to the sum of their structure sheaves, chi(O) also goes down by n-1.  
Thus both invariants are the same for degree one curves and change the same way when we raise degree, hence they agree.  But all curves are at least nodal plane curves, and we can modify the argument to allow for the nodes.
As to why there exist g holomorphic differentials, by taking real and imaginary parts this is equivalent to why there are 2g harmonic differentials.  I.e. we are trying to understand why Hodge’s theorem is true.  As mentioned, the maximum modulus principle implies a non zero harmonic form must have non zero period over some loop, and a harmonic form is uniquely determined by its periods over all homology cycles.  Thus existence of harmonic forms means finding ones with prescribed periods.  This is done in two steps.  The first step is to find a smooth form with given periods.  This is done by putting a collar around a non separating loop, forming a smooth function that climbs from 0 to 1 when crossing the collar, setting it equal to zero elsewhere, and taking d of it.  Thus we see that topology does tell us how many smooth closed forms we can construct that are cohomologically independent.  Then the deep part, i.e. the Hodge theory, says that each such smooth closed form can be written as a sum of a harmonic form and an exact form.  For details see Springer chapters 7 and 8 page 207, lemma 8-1, especially using Weyl’s smoothing lemma 7.3 p. 199.
To “see” why this occurs, one may view the flow diagrams in Springer’s chapter 1 (fig.1.27. p.27), taken from Klein’s lectures on Riemann surfaces.  I.e. it seems a differential form, or covector field, is determined by its flow lines, and when these flow lines are of minimal length in some sense, then they satisfy the Laplace equation and hence are harmonic.  So the intuitively plausible fact that these lines may be allowed to deform within a cohomology class until they are minimal (Dirichlet principle), suggests the existence of a harmonic form in each cohomology class (???  I just made this up, but maybe some expert will help out here.)
It seems Riemann reduced this problem to the “Dirichlet problem” of existence of a harmonic function with prescribed boundary conditions, on the polygon of 4g sides obtained by dissecting the genus g Riemann surface by 2g transverse cuts.  One constructs a harmonic function f which agrees on opposite pairs of sides except on one pair where it differs by a ≠ 0 constant on the two identified edges.  Then taking df gives a harmonic differential with a non trivial period.  Since there are 2g choices of such paired sides, and a harmonic form is determined by its periods, one can construct exactly 2g independent harmonic differentials this way, hence g holomorphic differentials.
A: To spell out potentially dense's comment, the Hodge theorem tells you that 
$$\begin{array}{rcl}
H^0(X, \mathbb{C}) &\cong& H^0(X, \mathcal{O}) \\
H^1(X, \mathbb{C}) &\cong& H^1(X, \mathcal{O}) \oplus H^0(X, \Omega^1) \\
H^2(X, \mathbb{C}) &\cong& H^1(X, \Omega^1) \\
\end{array}$$
where the left hand sides are topological cohomology and the right hand sides are sheaf cohomology.
Serre duality tells us that $H^1(X, \mathcal{O}) \cong H^0(X, \Omega^1)^{\vee}$ and $H^0(X, \mathcal{O}) \cong H^1(X, \Omega^1)^{\vee}$. So
$$\dim H^0(X, \mathbb{C}) - \dim H^1(X, \mathbb{C}) + \dim H^2(X, \mathbb{C}) = 2 \left( \dim H^0(X, \mathcal{O}) - \dim H^1(X, \mathcal{O}) \right)$$
as desired.
A: It is a consequence of the Hirzebruch-Riemann-Roch Theorem. 
In fact, for any line bundle $\mathscr{L}$ on a complete curve $X$ we have $$\chi(\mathscr{L}) = c_1(\mathscr{L}) + \frac{1}{2} \int_X c_1(T_X).$$
Taking $\mathscr{L}=\mathscr{O}_X$ and recalling that the integral over $X$ of the top Chern class of the tangent bundle is the topological Euler number $\chi_{\textrm{top}}(X)$, we obtain $$2 \chi(\mathscr{O}_X) = \chi_{\textrm{top}}(X).$$
