$\newcommand{\Cv}{\mathcal{C}}\DeclareMathOperator{\Atan}{atan2}$In the diagram, and as clarified in the comments, the curve $\Cv$ is a polar graph $r = f(\theta)$, for some positive function $f$ to be determined from the component functions $(x, y)$ of the parametrization. In terms of $f$, the desired distance function is
$$
d_{\theta} = f(\theta) + f(\theta + \pi).
$$
To find $f$, write $\theta = \Theta(t)$ and
\begin{alignat*}{2}
x(t) &= r\cos\theta &&= f(\theta) \cos\theta, \\
y(t) &= r\sin\theta &&= f(\theta) \sin\theta,
\end{alignat*}
so that
$$
f(\theta) = \sqrt{x(t)^{2} + y(t)^{2}}
\tag{1}
$$
and
$$
\Theta'(t) = \frac{d\theta}{dt} = \frac{y(t) x'(t) - x(t) y'(t)}{x(t)^{2} + y(t)^{2}}.
\tag{2}
$$
Under the geometric assumptions on $\Cv$, the preceding expression is non-vanishing, and without loss of generality (i.e., assuming $\Cv$ is traced counterclockwise) may be taken to be positive. If $[a, b]$ is the domain of the given parametrization of $\Cv$, then
$$
\Theta(t) = \int_{a}^{t} \frac{y(\tau) x'(\tau) - x(\tau) y'(\tau)}{x(\tau)^{2} + y(\tau)^{2}}\, d\tau
\tag{3}
$$
defines an invertible function $\Theta:[a, b] \to [0, 2\pi]$. The function $f$ is given by (1):
$$
f(\theta) = \sqrt{x(\Theta^{-1}(\theta))^{2} + y(\Theta^{-1}(\theta))^{2}}.
$$
While this analytic expression is exact, it entails inverting the definite integral (3), and may therefore be inconvenient for practical use. If instead the goal is to calculate $d_{\theta}$ for finitely many specified values of $\theta$, it suffices, for a given $\theta$, to find (using Newton's method, say) the values $t_{1}$ and $t_{2}$ such that
$$
\Atan(y(t_{1}), x(t_{1})) = \theta,\qquad
\Atan(y(t_{2}), x(t_{2})) = \theta + \pi,
$$
so that
$$
d_{\theta} = \sqrt{x(t_{1})^{2} + y(t_{1})^{2}} + \sqrt{x(t_{2})^{2} + y(t_{2})^{2}}.
$$
As usual, $\Atan(y, x)$ denotes the branch of polar angle of the ray from the origin through $(x, y)$ and taking values in $(-\pi, \pi)$.
