Why is the fundamental group of a compact Riemann surface not free ?

You can use covering spaces instead of cohomology.

(Edit: This proof does not use a presentation of the group, either -- just the weaker fact that its abelianization is free of rank $2g$.)

As Daniel pointed out in his comment, if the group were free it would have to be free on $2g$ generators. The genus $g$ surface has a $2$-sheeted covering space which is a genus $2g-1$ surface. Every index $2$ subgroup of a free group on $r$ generators is free on $2r-1$ generators, because it is the fundamental group of a $2$-sheeted covering space of a wedge of $r$ circles. So $2(2g-1)=2(2g)-1$, a contradiction.

As per Theo's request, I'm posting this as an answer, though it's largely an expansion on Vitali's comment. Let $F_n$ be the free group on $n$ letters; $K(F_n, 1)$ is a wedge of $n$ circles and so has vanishing cohomology in degrees $>1$. On the other hand, if $X$ is a compact Riemann surface of genus $g>1$, $X$ is a $K(\pi_1(X), 1)$ as its universal cover is the upper-half plane, which is contractible. But then $$H^2(\pi_1(X), \mathbb{Z})=H^2_{sing}(X, \mathbb{Z})=\mathbb{Z},$$ which is non-zero. In particular, $\pi_1(X)$ has non-vanishing cohomology in degree $2$ and so is not free, as Vitali says.

There are many good answers already, but maybe you'll be interested in a counting argument.

Let $Q_8$ denote the quaternion group. Consider the set

$$\mbox{Hom}(\pi_1(X),Q_8)$$

where $X$ is the closed surface of genus $g$. Since we have a presentation for $\pi_1(X)$, we note that a homomorphism to $Q_8$ is given by a $2g$-tuple of elements in $Q_8$ satisfying some group word. In particular, the cardinality of the set is finite. Indeed, if the fundamental group were free, the cardinality would be a power of $8$. However, combinatorics (or gentle representation theory) gives the cardinality of the set exactly:

$$|\mbox{Hom}(\pi_1(X),Q_8)| = 2^{6g-1}+2^{4g-1}$$

which (for $g>0$) is never a power of $2$, still less a power of $8$.

---Edit---

The "gentle representation theory" to which I refer can be found here: http://arxiv.org/abs/1102.4353

We show in section 4 that the cardinality of $\mbox{Hom}(\pi_1(X),G)$ can be written explicitly in terms of the dimensions of the irreducible representations of $G$.