We regard $D^2$ and its boundary circle $S^1$ as subspaces of $\mathbb C$.
Let $p : D^2 \to X = D^2/{\sim}$ be the quotient map. It maps the open unit disk $\mathring{D}^2$ homeomorphically onto $p(\mathring{D}^2)$. Thus we can regard $\mathring{D}^2$ as a genuine open subspace of $X$. Also note that the restrictions $p_0 = p \mid_{D^2 \setminus \{0\}}: D^2 \setminus \{0\} \to p(D^2 \setminus \{0\}) = X \setminus \{0\}$ and $p_1 = p \mid_{S^1} : S^1 \to p(S^1)$ are a quotient map because $D^2 \setminus \{0\}$ is a saturated open subset of $D^2$ and $S^1$ is a saturated closed subset of $D^2$ ($M$ saturated means $M = p^{-1}(p(M))$; now see The restriction of a quotient map to a saturated closed set is a quotient map).
The subspace $\bar S^1 = p(S^1) \subset X$ is a copy of $S^1$:
Consider the map $f : S^1 \to S^1, f(z) = z^n$. This is a continuous surjection with the property that $f(z) = f(z')$ iff $p_1(z) = p_1(z')$. Thus $f$ induces a continuous bijection $\bar f : \bar S^1 \to S^1$. Since $\bar S^1$ is compact and $S^1$ is Hausdorff, the map $\bar f$ is a homeomorphism.
Let $U = X \setminus \{0\}$ and $V = \mathring D^2$. Then $U, V$ are open subsets of $X$ such that $U \cup V = X$ and $U \cap V = \mathring D^2 \setminus \{0\}$. The embedding $\iota : S^1 \to \mathring D^2 \setminus \{0\}, \iota(z) = z/2$, is a homotpy equivalence. Moreover $\bar S^1$ is a strong deformation retract of $U$:
Define $H : (D^2 \setminus \{0\}) \times I \to D^2 \setminus \{0\}, H(z,t) = (1-t)z + tz/\lvert z \rvert$. This homotopy describes a strong deformation retraction of $D^2 \setminus \{0\}$ to $S^1$. It is easy to verify that $p_1H(z,t) = p_1H(z',t)$ iff $p_1(z) = p_1(z')$. Since $p_1 \times id_I$ is a quotient map by compactness of $I$, we get an induced homotopy $\bar H : U \times I \to U$. Clearly $\bar H$ describes a strong deformation retraction of $U$ to $\bar S^1$. In particular $r : U \to \bar S^1, r([z]) = \bar H([z],1)$, is a homotopy equivalence.
By construction the composition
$$S^1 \stackrel{\iota}{\to} \mathring D^2 \setminus \{0\} = U \cap V \hookrightarrow U \stackrel{r}{\to} \bar S^1 \stackrel{\bar f}{\to} S^1$$
is nothing else than $f$.
We have $\pi_1(U) = \pi_1(U \cap V) = \mathbb Z, \pi_1(V) = 0$. Seifert - van Kampen tells us that $\pi_1(X) \approx \left(\pi_1(U) * \pi_1(V)\right)/N$, where $N$ is the normal subgroup generated by all words of the form $i_U(g)i_V(g)^{-1}$, where $g \in \pi_1(U \cap V)$ and $i_U : \pi_1(U \cap V) \to \pi_1(U)$ and $i_V : \pi_1(U \cap V) \to \pi_1(V)$ are the inclusion-induced homomorphisms. Here we have $\pi_1(V) = 0$, thus
$\pi_1(X) \approx \pi_1(U)/N'$, where $N'$ is the normal subgroup generated by all $i_U(g)$. Since $\pi_1(U) \approx \mathbb Z$ is abelian, $N'$ is nothing else than the image of $i_U$. But with the homotopy equivalences $\iota : S^1 \to U \cap V$ and $\bar f r :U \to S^1$ we get a commutative diagram
$\require{AMScd}$
\begin{CD}
\pi_1(U \cap V) @>{i_U}>> \pi_1(U) \\
@A{\iota_*}A{\approx}A @V{\approx}V{(\bar f r)_*}V \\
\pi_1(S^1) @>{f_*}>> \pi_1(S^1) \end{CD}
Thus $\pi_1(X) \approx \pi_1(U)/\operatorname{im}(i_U) \approx \mathbb Z/\operatorname{im}(f_*) = \mathbb Z/n \mathbb Z$.
