Consider a curve $\gamma: [a,b] \to D \subset \mathbb{R}^2$, which is "piecewise-regular", that is:
$$\gamma(t)=\begin{cases} \gamma_1(t) & t \in[a=t_1,t_2] \\ \gamma_2(t) & t \in [t_2,t_3] \\ ... \\ \gamma_{N} (t) & t \in[t_N,b=t_{N+1}] \end{cases}$$
Where $a=t_1<t_2<...<t_N<t_{N+1}=b$ and $\gamma_i$ $\forall i =1,...,N+1$ are regular curves (which means differentiable where defined and with $\gamma' \neq \bar{0}$ everywhere).
Suppose also that $\gamma$ is closed, that is $\gamma(a)=\gamma(b)$.
Then let $r:D \subset \mathbb{R}^2 \to \mathbb{R}^3$ be a surface with $r \in C^2(D)$ (not necessarily regular).
Is the composition of the two $r \circ \gamma:[a,b] \to \mathbb{R}^3$ is still a "piecewise regular" curve?
No: For instance, a curve segment could trace a portion of the preimage of a point under $r$. As a specific example, consider the (geographer's convention) spherical coordinates parametrization $$ r(\theta, \phi) = (\cos\theta \cos\phi, \sin\theta \cos\phi, \sin\phi) $$ defined on $D = [-\pi, \pi] \times [-\pi/2, \pi/2]$, with $\theta$ representing longitude and $\phi$ latitude. The boundary of the rectangle can be trivially parametrized as a piecewise regular curve, but the top and bottom edges (i.e., the segments where $|\phi| = \pi/2$) map to the north and south poles, respectively.