Here is another example using only algebraic and not transcendental functions. Take three distinct complex numbers $z_1,z_2,z_3$ and let $X$ be the Riemann surface obtained by integrating the form $\sqrt{(z-z_1)(z-z_2)(z-z_3)}dz$. The map $g: X \to\mathbb C$ which sends each germ $\varphi_z\in X$ to $z$ is a covering with infinitely many sheets above the $z$-plane. (The ramification points are exactly above $z_1,z_2,z_3$, and there are infinitely many; each one is of order 2. This is so because near $z_k$ the form is like $(z-z_k)^{1/2}dz$ and therefore the primitive is like $(z-z_k)^{3/2}+cte$.) Now consider the holomorphic map $f: X\to\mathbb C$ which sends each germ $\varphi_z\in X$ to $\varphi_z(z)$. The branch points of $f$ are exactly the same as the branch points of $g$, but whilst they are of order 2 for $g$ they are of order 3 for $f$. (To see why this is so resolve the equation $\zeta=(z-z_k)^{3/2}+cte$ in $z$.) Now comes the important point: For nearly all choices of $z_1,z_2,z_3$ the periods $a_1,a_2,a_3$ of $ydz$ on the elliptic curve $y^2=(z-z_1)(z-z_2)(z-z_3)$ are such that the lattice $\mathbb{Z}a_1+\mathbb{Z}a_2+\mathbb{Z}a_3$ is dense in the plane. This means that the set of the images of the branch points of $f$ is dense in the plane.
In this way you can construct many other examples...