To understand why this is true I have been told to imagine the covering space as ..."drawing a circle wrapping around a cylinder $n$ times with $(n-1)$ intersections".
The covering space $p: \mathbb{R} \to S^1$ defined as $p(x)=(\cos 2 \pi x, \sin 2 \pi x)$ makes sense to me because all we are doing is finding all the values $x \in \mathbb{R}$ which map to the same point on $S^1$.
So, we can just draw a line and easily demonstrate these intervals. We do not need to imagine $\mathbb{R}$ as anything else.
I am confused as to why we are now imagining $S^1$ as something else then $S^1$.
I am sorry if my question is confusing. Anyways, I would just appreciate a clear answer as to why this is a covering space.
As has been suggested, it seems appropriate to get away from the confusing attempt at an intuitive picture and simply prove something. Thus let's forget all ideas about cylinders and springs.
For simplicity take $n=2$. Then the inverse image of a point $e^{i\theta}$ in $S^1$ is $\{e^{i\theta/2},e^{i(\theta/2+\pi)}\}$. In particular, inverse images are discrete. Let's show that $1$ is covered. Indeed, the inverse image of $S^1\setminus \{-1\}$ is the disjoint union of $\{e^{ i\theta}:\theta\in(-\pi/2,\pi/2)\}$ and $\{e^{ i\theta}:\theta\in(-3\pi/2,-\pi/2)\}$, both of which are mapped homeomorphically onto $S^1\setminus \{-1\}$. A similar argument works to show $z^n$ covers at different points and for different $n$.
Let me attempt, anyway, to clarify the intuitive picture you've been given: it should look like the following, where $p$ is the covering map and $L$ is some point in the base circle $S$. This picture does make it obvious for many people that $p$ is covering, since the inverse image of a small enough interval around $L$ is the union of two intervals on the locally-distinct components of the covering circle lying above $L$. You could draw it on the cylinder as you've been advised, but then you'd have to remember that an apparent intersection in the fiber over $L$ wasn't really there.