Update: I put my understanding in the answer.
If you find any mistakes, please let me know. Thanks for your time.
This is from Hatcher's Algebraic Topology, Example 2.43: Lens Spaces, page 144--146.
Given integer $m > 1$ and integers $l_{1}, \ldots, l_{n}$ relatively prime to $m$. Define action of $\mathbb Z_{m} = \langle \rho \rangle$ on $S^{2n-1} \subset \mathbb C^{n}$ by $\rho(z_{1}, \ldots, z_{n}) = (e^{2\pi i l_{1}/m} z_{1}, \ldots, e^{2\pi i l_{n}/m} z_{n})$.
Lens space $L = L_{m}(l_{1}, \ldots, l_{n}) := S^{2n-1} / \mathbb Z_{m}$.
Devide unit circle $C$ in $n$-th $\mathbb C$ factor of $\mathbb C^{n}$ by taking points $e^{2\pi ij/m} \in C$ as vertices, $j = 1, \ldots, m$.
Joining the $j$-th vertex of $C$ to unit sphere $S^{2n-3} \subset \mathbb C^{n-1}$ by arcs of great circles in $S^{2n-1}$.
All these arcs form a $(2n-2)$-dimensional ball $B_{j}^{2n-2}$, which is homeomorphic to $D^{2n-2}$ and bounded by $S^{2n-3}$.
Similarly, joining the $j$-th edge of $C$ to $S^{2n-3}$ gives a ball $B_{j}^{2n-1}$ bounded by $B_{j}^{2n-2}$ and $B_{j+1}^{2n-2}$.
Rotation $\rho$ takes $S^{2n-3}$ to itself while rotating $C$ by angle $2\pi l_{n} / m$,
hence $\rho$ permutes the $B_{j}^{2n-1}$'s and $B_{j}^{2n-2}$'s.
For $r$ s.t. $r l_{n} = 1 \bmod m$, $\rho^{r}$ takes each $B_{j}^{2n-1}$ and $B_{j}^{2n-2}$ to the next one.
$(r, m) = 1$, $\rho^{r}$ is a generator of $\mathbb Z_{m}$.
We may obtain $L$ as quotient of one $B_{j}^{2n-1}$ by identifying its two faces $B_{j}^{2n-2}$ and $B_{j+1}^{2n-2}$ together via $\rho^r$.
The two faces $B_{j}^{2n-2}$ and $B_{j+1}^{2n-2}$ are identified via a reflection across $B_{j}^{2n-1}$ fixing $S^{2n-3}$, followed by a rotation.
My question:
$1$. What's the relation of these $B_{j}^{2n-1}$'s? Do we have $\bigsqcup_j B_{j}^{2n-1} = S^{2n-1}$?
$2$. Since $B_{j}^{2n-2}$ and $B_{j+1}^{2n-2}$ identified together via $\rho^r$, where does that reflection come from?
In the first picture, I visualize $S^{2n-3}$ as $S^0$ (I really doubt the legitmacy of this, since $S^0$ is not connected and other $S^n$'s are connected), visualize $S^{2n-1}$ as $S^2$, $B_j^{2n-2}$ and $B_{j+1}^{2n-2}$ as arcs in $S^{2}$ connecting "north pole" and "south pole" along great ciecle, and region between them is $B_j^{2n-1}$ in the second figure.
In the third picture, I visualize $S^{2n-3}$ as $S^1$, $B_j^{2n-2}$ and $B_{j+1}^{2n-2}$ as upper and lower disk $D^{2}_+$ and $D^{2}_-$.
Arcs in red, green, yellow and purple are identified respectively.
As far as I understand right now, the unit circle $C$ in the $n$-th $\mathbb C$ factor is vertical to $S^{2n-3}$, so arcs along graet circles in $S^{2n-1}$ from points $e^{2\pi ij/m}$ $(j=1,\cdots, m)$ to points in $S^{2n-3}$ yield $(2n-2)$-dimensional balls $B_{j}^{2n-1}$, which form an equal division of $S^{2n-1}$.
$\rho^r$ is a generator of $\mathbb Z_m$ and takes each $B^{2n−1}_j$ and $B^{2n−2}_j$ to the next one, so we can obtain $L$ as quotient of only one $B^{2n−1}_j$ by identifying its two faces $B^{2n−2}_j$ and $B^{2n−1}_{j+1}$ via $ρ^r$.
Since $\rho$ acts on different coordinates respectively, we can factorize $\rho^{r}$ as $\rho_{2} \circ \rho_{1}$.
$\rho_{1}$ is $\rho^{r}$ restrcting on last $\mathbb C_{n}$ factor, producing a reflection from $B_{j}^{2n-2}$ to $B_{j+1}^{2n-1}$ fixing $S^{2n-3}$.
It's indeed a reflection since the unit circle $C$ in the $n$-th $\mathbb C$ factor is vertical to $S^{2n-3}$, and $B_{j}^{2n-2}$ is formed by arcs along great circles in $S^{2n-1}$.
$\rho_{2}$ is $\rho^{r}$ restrcting on $S^{2n-3}$, producing a rotation on $S^{2n-3}$.