Let $C$ be a smooth complex projective curve of genus $g$ and let $C^{(2)}$ be the symmetric product of $C$ with itself, i.e. the quotient of $C\times C$ by the involution $\sigma$ which exchanges the two coordinates. To compute the second Betti number of $C^{(2)}$ I am reasoning as follows. First of all it should be $$H^2(C^{(2)}) = H^2(C\times C)^\sigma$$ (maybe here is the mistake?). Then, using Künneth theorem, it follows that $$\dim H^2(C\times C)^\sigma = 1 + \frac{2g(2g+1)}{2},$$ the first summand is the dimension of the diagonal of $H^2\otimes H^0\oplus H^0\otimes H^2$, while the second summand is the dimension of symmetric tensors in $H^1\otimes H^1$.
If $g=2$ the above formula gives dimension $11$. On the other hand in this case $C^{(2)}\to JC$, defined by $p+q\mapsto [p+q-K_C]$ is a birational morphism which contracts a unique excepcional divisor, the graph of the unique $g_2^1$ of $C$, so that the dimension is $1$ more than the second Betti number of $JC$ which should be $6$ (again by Künneth).
Could you please point out the mistake?