Funny how this site works sometimes.
I was doing a little research on Desmos for this completely unrelated question when I realized that the method of restricting domain and range I ended up using to answer that question would also work perfectly here. I had to switch to the rectangular versions of the petal equations, but that's a small price to pay.
Just as a quick example, I ended up graphing
$$x^2+y^2=1.21$$
$$\begin{align}
\frac{(x\cos\phi_1-y\sin\phi_1-h_1)^2}{a_1^2}+\frac{(y\cos\phi_1+x\sin\phi_1-k_1)^2}{b_1^2}&=c_1^2\{x\le-0.5\}\{y\ge0.96465\} \\
\frac{(x\cos\phi_1-y\sin\phi_1-h_1)^2}{a_1^2}+\frac{(y\cos\phi_1+x\sin\phi_1-k_1)^2}{b_1^2}&=c_1^2\{x\ge-0.49999\}\{y\ge1.09571\} \\
\frac{(x\cos\phi_2-y\sin\phi_2-h_2)^2}{a_2^2}+\frac{(y\cos\phi_2+x\sin\phi_2-k_2)^2}{b_2^2}&=c_2^2\{x\le-1.29445\} \\
\frac{(x\cos\phi_2-y\sin\phi_2-h_2)^2}{a_2^2}+\frac{(y\cos\phi_2+x\sin\phi_2-k_2)^2}{b_2^2}&=c_2^2\{x\le-0.99377\}\{y\le0.47522\} \\
\frac{(x\cos\phi_2-y\sin\phi_2-h_2)^2}{a_2^2}+\frac{(y\cos\phi_2+x\sin\phi_2-k_2)^2}{b_2^2}&=c_2^2\{x\ge-0.72153\}\{y\ge0.84065\}
\end{align}$$
to obtain the center and the first two petals, and the result looked exactly like what I had hoped for:
I wouldn't even dream of typing out all $40$ equations I ended up graphing, but here's the final result, which looks considerably nicer than what I started with:
And here's the link to the final version, in case anyone's curious.
Your cone is defined by
$$
\{x^2+y^2=k^2z^2\}\,.
$$
A possible parametrization using 2D polar coordinates for $\mathbb R^2$
and cylindrical coordinates for $\mathbb R^3$ is:
$$
i:(r,\theta)\mapsto (r,\theta,ar)\,,\text{ where }a:=\frac{1}{k}\,.
$$
(Compared to the pictures below the cone is inverted with tip at $(0,0,0)$.
This trivial "$z$-flip" does not change the statements about its geodesics.)
The Euclidean metric of $\mathbb R^3$ in cylindrical coordinates is
$$
ds^2=dr^2+r^2\,d\theta^2+dz^2\,.
$$
Since $dz=a\,dr$ this leads to
$$\tag{1}
ds^2=(1+a^2)\,dr^2+r^2\,d\theta^2
$$
for the cone.
The cone metric (1) becomes
$$\tag{2}
ds^2=d\rho^2+\rho^2\,d\varphi^2
$$
under the transformation
$$\tag{3}
r\mapsto \rho:=\sqrt{1+a^2}\,r \,,\quad\theta\mapsto \varphi:=\frac{\theta}{\sqrt{1+a^2}}\,.
$$
The metric (2) is formally equal to the Euclidean metric of $\mathbb R^2$ when
the radius is stretched by the factor $\sqrt{1+a^2}$ and the angle is reduced by the factor $1/\sqrt{1+a^2}\,.$ The pictures below should make this self explanatory.
Note that $a$ is nothing else than
$$
a=\frac{z}{r}=\cot\alpha
$$
where $\alpha$ is half of the angle at the tip of the cone. Therefore
$$
\sqrt{1+a^2}=\sin\alpha\,.
$$
The range of $\theta$ is always $[0,2\pi)\,.$ The range of $\varphi$ is less than that which is clear from looking at the flattened (unfolded) cone in the
$(\rho,\varphi)$-plane.
Knowing that the Euclidean metric in polar coordinates has only straight lines
as geodesics it becomes clear from (2) that the geodesics on the cone are straight lines on the unfolded cone in the $(\rho,\varphi)$-plane.
The following pictures were generated with GeoGebra. They show that geodesics on the cone can intersect with themselves. Nonetheless they are still straight lines
after unfolding.
Appendix: Clairaut's Theorem
For the $(\rho,\varphi)$-plane let's introduce "Cartesian" coordinates
$$
\xi=\rho\cos\varphi\,,\quad \eta=\rho\sin\varphi\,.
$$
A geodesic being a straight line means that there is a parametrization of $\rho,\varphi$ with $t$ such that the components of its tangent,
$$
\dot\xi=\dot\rho\cos\varphi-\rho\,\dot\varphi\,\sin\varphi\,,\quad\dot\eta=\dot\rho\sin\varphi+\rho\,\dot\varphi\,\cos\varphi,
$$
are constant.
A latitude on the cone is a circle in the
$(\rho,\varphi)$-plane. Its components are
$$
\lambda=c\cos\varphi\,,\quad\mu=c\sin\varphi
$$
with a constant $c>0\,.$ Up to that constant,
the scalar product of the tangent at the geodesic and at the latitude
is
\begin{align}\require{cancel}
\dot\xi\dot\lambda+\dot\eta\dot\mu&=
-\cancel{\dot\rho\,\dot\varphi\,\cos\varphi\,\sin\varphi}+\rho\,\dot\varphi^2\,\sin^2\varphi+\cancel{\dot\rho\,\dot\varphi\,\sin\varphi\cos\varphi}+\rho\,\dot\varphi^2\,\cos^2\varphi\\
&=\rho\,\dot\varphi^2\,.
\end{align}
Since the length of the tangent at the latitude is $c\dot\varphi$ and
the length of the tangent at the geodesic is the constant
\begin{align}
\sqrt{
\dot\xi^2+\dot\eta^2}&=\sqrt{\dot\rho^2+\rho^2\,\dot\varphi^2}
\end{align}
it follows that the cosine of the angle between geodesic and latitude
is
$$
c\frac{\rho\,\dot\varphi}{\sqrt{\dot\rho^2+\rho^2\,\dot\varphi^2}}\,.
$$
A general theorem by Clairaut about geodesics on surfaces of revolution states that $\rho$ times this cosine is constant.
Since the denominator is constant. Clairaut's theorem states in
our case that
\begin{align}\tag{4}\boxed{\phantom{\Bigg|}\quad
\rho^2\,\dot\varphi\quad\text{ is constant. }\quad}
\end{align}
This is indeed the case. Taking the derivative
yields $2\rho\,\dot\rho+\rho^2\,\ddot\varphi\,.$
Calculating a few Christoffel symbols it is not
difficult to see that the metric (2) leads to the geodesic equations
$$
0=\ddot \rho-\rho\,\dot\varphi ^2\,,\quad
0=\ddot \varphi+\frac{2}{\rho}\,\dot \rho\,\dot\varphi\,.
$$
So Clairaut's statement in this case is just the second of those equations.
By the simple transformation (3) Clairaut's statement carries over to
the cone in is cylindrical coordinates $(r,\theta,z)$ from above.
Best Answer
In a first round of $2\pi$ the parameter $t$ goes from $0$ to $0.5$ and the radius increase from $0.2$ to $0.6$, so \begin{align} x = \left(0.2+\frac{0.6-0.2}{0.5}t\right)\sin(4\pi t),\\ y = \left(0.2+\frac{0.6-0.2}{0.5}t\right)\cos(4\pi t),\\ \end{align} with $t\in[0,1].$