Here is a solution which, while not as elegant as the one using an electrostatic potential, is more elementary.
Since you already understood the cases $n=1$ and $2$, I will assume that $n\ge 3$.
First of all, I will fix the standard (counter-clockwise) orientation on the circle.
For each tuple $z=(z_1,...,z_n)\in X_n=Conf(S^1)$ the orientation on $S^1$ determines a cyclic order $o_z$ on the set of labels $[n]=\{1,...,n\}$.
Let ${\mathcal O}$ denote the set of all cyclic orders on $[n]$.
If $o_{z}\ne o_{z'}$ then, clearly $z, z'$ belong to distinct components of $X_n$. For each cyclic order $o$ on $[n]$, let $X_n^o$ denote the subset of $X_n$ consisting of configurations $z$ such that $o_z=o$. The permutation group $\Sigma_n$ acts naturally on $X_n=Conf(S^1)$ permuting the collection
$$
\{X_n^o: o\in {\mathcal O}\}
$$
It is clear that the action is transitive with the stabilizer of $X_n^o$ isomorphic to the cyclic subgroup of $\Sigma_n$ generated by the permutation $(2,3,...,n,1)$.
Therefore, it suffices to understand the topology of $X_n^o$ for one cyclic order $o$, the one which agrees with the orientation on the circle. Therefore, from now on, I will consider only $X_n^o$. There is one more group action which will help us, namely, $G=PSL(2,R)$ acting on the circle by linear-fractional transformations. This group acts of $X_n$ by
$$
(z_1,...,z_n)\mapsto (gz_1,...,gz_n), g\in G.
$$
This action preserves $X_n^o$. I will leave it to you to check that the action is proper and free, hence,
$$
X^o_n\to B=X^o_n/G
$$
is a principal $G$-bundle. I will prove that the base $B$ is contractible, actually, is homeomorphic to ${\mathbb R}^{n-3}$, from which it will follow that $X_n^o$ is homotopy-equivalent to $G$, i.e. to $S^1$. In fact,
$X^o_n$ is diffeomorphic to $G\times B$ by contractibility of the base.
In order to understand $B$, I will use a slice (a cross-section) for the action of $G$ on $X_n^o$: It will be a properly embedded submanifold $S\subset X^n_o$ such that:
a. The $G$-orbit of $S$ is the entire $X_n^o$.
b. For every $g\in G-\{1\}$, $g(S)\cap S=\emptyset$.
From this it will follow that the restriction to $S$ of the projection $X^o_n\to B$ is a diffeomorphism $S\to B$.
The construction of $S$ is quite simple: Pick three distinct points on $S^1$, say, $1, i, -1$, and define
$$
S=\{z=(z_1,...,z_{n-3},1,i,-1)\in X^o_n\}.
$$
Since the group $G$ acts simply-transitively on $X^o_3$ (the set of ordered triples of distinct points on $S^1$ with the standard cyclic orientation), $S$ indeed forms a slice of the action of $G$ on $X^o_n$. Thus, it remains to understand the topology of $S$. If $n=3$ then $S$ is a singleton, so we are done. I will assume now that $n\ge 4$.
Since the cyclic orientation $o$ is the standard one,
for all $z\in S$, the points $z_1,...,z_{n-3}$ all lie in the lower half-plane on the arc between $-1$ and $1$ and their projections $x_1,...,x_{n-3}$ (their real parts if you like complex numbers) to the horizontal axis satisfy
$$
-1< x_1<x_2<...<x_{n-3}<1.
$$
Define the map
$$
h: S\to \Delta=\{(x_1,...,x_{n-3})\in {\mathbb R}^{n-3}:
-1 <x_1<...<x_{n-3}<1,$$
$$
h: z=(z_1,...,z_{n-3},1,i,-1)\mapsto (Re(z_1),..., Re(z_{n-3})),
$$
where I am thinking of each $z_k$ as a complex number.
I will leave it to you to verify that the map $h$ is a diffeomorphism. (The "homeomorphism" part is quite easy and it suffices for your purposes.) Lastly, $\Delta$ is an open nonempty convex subset in ${\mathbb R}^{n-3}$, hence, is diffeomorphic to ${\mathbb R}^{n-3}$. Thus, we obtain:
$$
X_n^o= PSL(2, {\mathbb R})\times {\mathbb R}^{n-3}.
$$
Let me know if you need help proving that $G=PSL(2, {\mathbb R})$ is homotopy-equivalent to $S^1$, or, better is diffeomorphic to ${\mathbb R}^2\times S^1$. One proof is to construct a diffeomorphism of $G$ to the unit tangent bundle of the hyperbolic plane: There is a natural simply-transitive action of $G$ on this bundle.
