How to think about Fundamental Groups I want to make sure I have the concepts of the Fundamental Groups well. The underlying set is that of the quotient set defined by the loops in the topological space, and the equivalence relation of path homotopy between said loops. Then, the operation of the group would be composing, by let's say two loops have base point $p$, then you get like a figure $8$ from composing the loops, and then you allow the figure $8$ to morph into a $0$, still with base point $p$, so that if the Fundamental Group has more elements, you can keep composing this way?
Next, my other question is about the fundamental group of the circle. Normally I would picture a path from $p$ to $p$ in a circle as unique. But from my understanding of the FG of the circle, a path from $p$ to $p$ that has passed $n$ times over $p$ is different from a path that has passed $m$ times over $p$ when $m\neq n$?
And the only reason you can't do this in a disk, is if supposedly you had a winded loop on the boundary of the unit disk, then you could actually unfold it within the disk, getting to the $n=1$ loop?
Are these thoughts the correct way of thinking about the fundamental group?
Edit: I'm partly asking this because the picture of composition in Wikipedia confused me:
https://upload.wikimedia.org/wikipedia/commons/thumb/4/4b/Homotopy_group_addition.svg/240px-Homotopy_group_addition.svg.png
Shouldn't you start with picture 2, then go to 1, and I don't know what 3 stands for.
Thanks in advance.
 A: Remember that we have to be concerned with more than just the image of a loop. 
A loop is a continuous function $f:[0, 1] \rightarrow X$ with $f(0)=f(1)=p$. 
Consider the constant loop $f:[0,1]\rightarrow\mathbb{C}$ defined by  $f(x)=1$ for all $x\in [0,1]$. This is certainly a different loop than the one which traverses halfway around the circle and then comes back: $g(x)=\begin{cases}
e^{2\pi ix} & x\in\left[0,\frac{1}{2}\right]\\
e^{2\pi i\left(1-x\right)} & x\in\left[\frac{1}{2},1\right]
\end{cases}$.
But $f$ and $g$ are homotopic through the homotopy $H:[0,1]\times[0,1]\rightarrow\mathbb{C}$ defined by $H(x, t) = \begin{cases}
e^{2\pi ixt} & x\in\left[0,\frac{1}{2}\right]\\
e^{2\pi i\left(1-x\right)t} & x\in\left[\frac{1}{2},1\right]
\end{cases}$, so they are in the same equivalence class and represent the same element in $\pi_1(S^1)$. 
It should be easy to see that similarly $f$ is also homotopic to a loop which traverses around the circle twice and then comes back. This loop "passes" the basepoint more than once but is homotopic to the constant map. So the number of times that the basepoint is in the image of the loop is not enough to determine the equivalence class of the loop.
You are correct in stating that any loop which traverses around the circle exactly $n$ times will not be homotopic to a loop which traverses it exactly $m$ times, with $n\neq m$ (where the loop is not allowed to "backtrack" like the previous examples) , but this takes some work to show.
I think I misinterpreted your question but still posting this in case it is helpful at all.
