Let $X=\{y^2=x^3-1\}$ in $\mathbb{C}^2$. We use the projection to the $x$ plane to analyse the topology of $X$. This projection is of degree $2$ and ramified over the cube roots of unity, $1,\eta,\eta^2$. Denote by $p=(-1/2,0)$ the midpoint between $\eta$ and $\eta^2$, any by $r$ the distance from $p$ to $\eta$ (or equivalently, the distance from $p$ to $\eta^2$).
Consider the loops $$\gamma_1(t)=(x(t),y(t))=\left(p+(r-1/3)e^{4\pi i t}, \sqrt{x(t)^3-1}\right),$$ $$\gamma_2(t)=(x(t),y(t))=\left(p+(r-1/2)e^{4\pi i t}, \sqrt{x(t)^3-1}\right).$$
Denote $\omega=3ydx-2xdy$. Then $$\int_{\gamma_1}\omega\not=0, \quad \int_{\gamma_2}\omega=0,$$ which I computed numerically using Mathematica. This means that the two loops cannot be homotopic. But when I try to visualise the Riemann surface by making branch cuts etc then I do not understand why these two loops are not homotopic.
I always thought that the homotopy type of a loop can be read of from the winding numbers that $x(t)$ makes around each of the ramification points, but for these two loops these winding numbers are all equal (and all 0), so this must be a misunderstanding on my part.
If you meant $x_1(t)=-\frac12+(1-a)(r-1/3)e^{4\pi i t},\gamma_1(t) = (x_{1}(t), \sqrt{x_{1}(t)^3-1})$ with $r=\frac{\sqrt3}2$ then it is homotopic to a point, and so is the other curve, with a branch of $\sqrt{x^3-1}$ analytic on $|x+\frac12|< \frac{\sqrt3}2$.