When considering the covering spaces of $S^1$ (I use as base point $0$), usually in textbooks or courses, you see the spiral with some winding number $n$ to represent a $n$-sheeted covering space. To pass to a universal cover, the explained "idea" is then to pull the ends of an infinitly long spiral such that it transforms to $\mathbb{R}$.
Now, I wanted to develop a similar inutition for the Riemann sphere $S^2$ (I use as base point also $0$) which by the uniformization theorem has a universal cover given by the Riemann sphere iself. So this intuition of pulling something infinitely long to unfold it, does not really work anymore. But also for the $n$-sheeted covering space case, I am stuck.
Taking into account that the fundamental group $\pi_1$ of $S^2$ is trivial and a covering $p$ has fundamental group isomorphic to $p_{\ast}(\pi_1)$, which is a subgroup of $\pi_1$ by the Unique Path Lifting Property and the index $|\pi_1:p_{\ast}(\pi_1)|=1$ gives the number of sheets, I come to the understanding that any covering space of $S^2$ has a single sheet and, in turn, is simply $S^2$. This seems somewhat strange, since it can be extended to any simply connected Riemann surface. Is this an obstruction to the construction of a covering space of a simply connected Riemann surface, which has more than $1$ sheet?
For the spaces that you describe, and generally for connected and locally simply connected spaces, there exists a Galois Correspondence $$ \{\text{Subgroups $H$ of $\pi_1(X)$}\} \stackrel{1:1}{\longleftrightarrow} \{\text{Covering maps $p_{H}\colon X_H\to X$}\}. $$ The details of this are found in any introductory algebraic topology textbook, I personally like the treatment in these Lecture notes by Clara Löh, found in chapter 2.3.5.
In particular, if the fundamental group of $X$ is trivial, there is only one path connected covering space map to $X$, the identity.