Integrating differential forms over homology classes gives an embedding $H^1(X,\mathbb Z) \hookrightarrow H^0(X,\Omega^1)^*,$ and the Jacobian is the quotient $H^0(X,\Omega^1)^*/H^1(X,\mathbb Z)$.
Note that $H^0(X,\Omega^1)$, and hence also its dual, is a complex vector space of dimension $g$, and so this expresses the Jacobian of the curve as $\mathbb C^g$ modulo the period lattice. If you want to actually realize it as a projective variety, the standard method is via $\theta$-functions.
How much more explicit do you want to get?
Added in response to the comment below:
For a hyperelliptic curve you can write down a basis of holomorphic differential (generalizing $dx/y$ in the elliptic curve case), but the lattice will (so to speak) be whatever it is. E.g. in the elliptic curve case, the lattice is (after rescaling) equal to the span of $1$ and $\tau$, where $\tau$ can be any element of the upper half-plane. In what sense do you want to identify $\tau$?
Perhaps one way is that in the elliptic curve case we can compute $\tau$ via certain hypergeometric functions in terms of the $\lambda$-invariant of the elliptic curve. Is this what you have in mind?
Added in response to the follow-up comment:
The curve $y^2 = x^6 - 1$ admits a surjection onto the curve $y^2 = x^3 - 1$, via $(x,y) \mapsto (x^2,y)$. Thus the latter curve (an elliptic curve with CM by $\sqrt{-3}$) is a factor (up to isogeny) of the Jacobian of $y^2 = x^6 - 1$. Thus this Jacobian is isogenous to the product of a pair of elliptic curves.