Reducing the nlab-theorem to tom Dieck's theorem breaks down when one tries to show that the interiors of $\tilde X_0, \tilde X_1$ cover $\tilde X$. At least there is no simple proof - but nevertheless it could be true. Anyway, we do not need it. In fact, tom Dieck's theorem relies on two ingredients:
Theorem (2.6.1) which states a pushout property for fundamental groupoids under the assumption that $X_0$ and $X_1$ are subspaces of $X$ such that the
interiors cover $X$.
The existence of a retraction functor $r : \Pi(Z) \to \Pi(Z,z)$ which tom Dieck only defines for path connected $Z$. This works as follows: For each object $x$ of $\Pi(Z)$ (i.e. each point $x \in Z$) we define $r(x) = z$. For the morphisms we proceed as follows: We choose any morphism $u_x : x \to z$ if $x \ne z$ and take $u_z = id_z$= path homotopy class of the constant path at $z$. Given a morphism $\alpha : x \to y$ in $\Pi(Z)$, we define $r(\alpha) = u_y \alpha u_x^{-1}$.
We shall see that 2. can be generalized so that we can prove
Theorem (Seifert - van Kampen). Let $X$ be a topological space and $X_0,X_1\subset X$ be subsets whose interiors cover $X$ such that $X_{01}=X_0\cap X_1$ is path connected. Then for any choice of base point $\ast\in X_{01}$
\begin{matrix}\pi_1(X_{01},\ast) & \to & \pi_1(X_0,\ast)\\
\downarrow&&\downarrow \\
\pi_1(X_1,\ast) & \to & \pi_1(X,\ast)
\end{matrix}
is a pushout in the category of groups.
Proof. As Tyrone suggested in his comment, for a pointed space $(Z,z)$ let us denote by $\tilde Z$ the path-component of $Z$ containing the basepoint $z$. From Jackson's and your answers we know that for $X_{01}$ path connected and $* \in X_{01}$ we have $\tilde X_{01} := \tilde X_0 \cap \tilde X_1 = X_{01}$ and $\tilde X = \tilde X_0 \cup \tilde X_1$. Note that $X_{01}$ path connected is essential for both equations.
We apply tom Dieck's retraction contruction to $Z = \tilde X$ and $z = * \in X_{01} = \tilde X_{01}$ by first chosing $u_x$ in $\Pi(X_{01})$ for all $x \in X_{01}$ (where of course $u_* = id_*$), then $u_x$ in $\Pi(\tilde X_0)$ for all $x \in \tilde X_0 \setminus X_{01}$ and finally $u_x$ in $\Pi(\tilde X_1)$ for all $x \in \tilde X_1 \setminus X_{01}$. Since $\Pi(X_{01}),\Pi(\tilde X_0), \Pi(\tilde X_1)$ are subcategories of $\Pi(\tilde X)$, this gives us a choice of $u_x$ in $\Pi(\tilde X)$ for all $x \in \tilde X$ providing a retraction $\tilde r : \Pi(\tilde X) \to \Pi(\tilde X,*) = \Pi(X,*)$. We extend it to a retraction $r : \Pi(X) \to \Pi(X,*)$ as follows: Given a morphisms $\alpha : x \to y$ in $\Pi(x)$, then $x,y$ belong to same path component $P$ of $X$. If $P = \tilde X$, we define $r(\alpha) = \tilde r(\alpha)$. If $P \ne \tilde X$, we define $r(\alpha) =id_*$. Consider the restriction $r_{01}: \Pi(X_{01}) \to \Pi(X,*)$. The category $\Pi(X_{01})$ is a subcategory of $\Pi(\tilde X,*)$ and by construction $r_{01}(\Pi(X_{01})) = \tilde r(\Pi(X_{01})) \subset \Pi(X_{01},*)$, i.e. we may regard $r_{01}$ as a map $r_{01} : \Pi(X_{01}) \to \Pi(X_{01},*)$. Next consider the restriction $r_0 : \Pi(X_0) \to \Pi(X,*)$. For the subcategory $\Pi(\tilde X_0) \subset \Pi(X_0)$ we have by construction $r_0(\Pi(\tilde X_0)) = \tilde r(\Pi(\tilde X_0)) \subset \Pi(X_0,*)$. Let $\tilde P_0$ be a path component of $X_0$ different from $\tilde X_0$, i.e. $\tilde P_0 \cap \tilde X_0 = \emptyset$. Then $\tilde P_0 \cap \tilde X = \tilde P_0 \cap \tilde X_1 = \tilde P_0 \cap X_0 \cap \tilde X_1 \subset \tilde P_0 \cap X_0 \cap X_1 = \tilde P_0 \cap \tilde X_0 \cap \tilde X_1 \subset \tilde P_0 \cap \tilde X_0 = \emptyset$. Thus $\tilde P_0 \cap \tilde X = \emptyset$ and therefore by construction $r_0(\Pi(\tilde P_0)) = r(\Pi(\tilde P_0) = \{id_*\} \subset \Pi(X_0,*)$. We conclude $r_0(\Pi(X_0)) \subset \Pi(X_0,*)$, i.e. we may regard $r_0$ as a map $r_0 : \Pi(X_0) \to \Pi(X_0,*)$. Similarly $r$ restricts to $r_1 : \Pi(X_1) \to \Pi(X_1,*)$. Therefore we get a commutative diagram
\begin{matrix}\Pi(X_0) & \hookleftarrow & \Pi_1(X_{01}) & \hookrightarrow & \Pi(X_1)\\
\downarrow r_0 && \downarrow r_{01} &&\downarrow r_1 \\
\Pi(X_0,*) & \hookleftarrow & \Pi(X_{01},*) & \hookrightarrow & \Pi(X_1,*)
\end{matrix}
Now the same argument as in the proof of tom Dieck's Theorem (2.6.2) applies.
Best Answer
Suppose we have a closed disc $\bar{D}^2$ whose boundary $\partial D^2$ is attached to a circle $S^1$ by a map $\gamma:\partial D^2\to S^1$ that wraps $\partial D^2$ in total $n$ times around $S^1$. We call the resulting space $Y_n$.
Let $U$ and $V$ be open subsets of $Y_n$ defined as follows. The set $U$ is given by $(U\cap D^2)\cup S^1$, where $D^2$ is the interior of $\bar{D}^2$, and $U\cap D^2$ is a narrow strip on the outer edge of $D^2$ (so that $\partial(U\cup D^2)$ contains $\partial D^2$). The set $V$ is just $D^2$.
My awful picture may help explain this. The space $Y_n$ (on the left, where the orange arrows denote the attaching map $\gamma$) is the union of $U$ (the yellow subset) and $V$ (the pink subset).
Note that each of $U$ and $U\cap V$ has a deformation retract to just $S^1$, but $V$ is contractible. That is $\pi_1(U)\cong\Bbb Z$, $\pi_1(V)\cong\{1\}$, and $\pi_1(U\cap V)\cong \Bbb Z$. Now observe that $\pi_1(U\cap V)\to \pi_1(U)$ is given by multiplication by $n$ because each simple loop in $U\cap V$ wraps $n$ times around $S^1$ (and $\pi(U\cap V)\to \pi_1(V)$ is trivial). By van Kampen's theorem, $$\pi_1(Y_n)=\pi_1(U)\underset{\pi_1(U\cap V)}{*}\pi_1(V)\cong (\Bbb Z*\{1\})/(n\Bbb Z)\cong \Bbb Z/n\Bbb Z.$$ Indeed, we can see that $\pi_1(Y_n)$ is generated by a generator $g$ of $\pi_1(S^1)$. When you have a loop homotopic to $ng$, it is homotopic to $\partial D^2$, and can then be contracted along $\bar{D}^2$ to a point.