First, notice that by reparameterizing the Grandi roses by replacing $t$ with $2 t$ (so, doubling the speed), we can write the curves respectively as graphs of polar functions $\rho, \textrm{P}$ in the angular variable (which I'll still call $t$):
\begin{align}
\rho &:= 1 + r \sin k t \\
\textrm{P} &:= R \sin k t .
\end{align}
(1) This viewpoint quickly explains both of the differences we notice in the plots with $k$ even: First, we see that if we plot both $\rho, \textrm{P}$ over a full period (an angular interval of $2 \pi$), the asymmetry of the plots with $k$ even disappears: With the original parameterization, we only trace for half a period of $\textrm{P}$, which is anyway less natural. Plotting both curves over a full period for even $k$ gives this more symmetric graph (for $r = 1 / 3, k = 6$).
We can also see immediately which this issue didn't occur for odd $k$, even with the slower parameterization. Expanding using the usual sum formula gives
$$\textrm{P}(t + \pi) = R \sin k(t + \pi) = (-1)^k R \sin k t = (-1)^k \textrm{P}(t).$$ But the polar coordinates $(t + \pi, \alpha)$ and $(t, -\alpha)$ represent the same point, which (for $r > 0$) tells us exactly that the parameterization $\textrm{P}(t)$ traces the graph of $\textrm{P}$ with period $\pi$ iff $k$ is odd. So, for $k$ odd we still get the complete graph by plotting it over an angular interval of length $\pi$, or equivalently over an angular interval of length $2 \pi$ (rather than $4 \pi$) in the original parameterization of the Grandi rose. The same identity also immediately more-or-less explains the more essential difference between the situations for $k$ even and odd, namely that for $k$ even (but not odd) the lobes of the curve $\textrm{P}$ extend outside the graph of $\rho$.
(2) We can see that for $r = 1$, $\rho$ has a minimum of zero---and so its graph intersects the origin---at integer multiples of $\frac{\pi}{k}$. When $r < 1$, $\rho(t) \geq 1 - r > 0$, in which case this behavior doesn't occur.
(3) It's really not clear to me what is meant here. But NB for $r < 1, k > 1$, the polar graphs of $\rho, \textrm{P}$ (as subsets of $\Bbb R^2$) are topologically inequivalent: The polar graph of $\rho$ is topologically equivalent to a circle, whereas the polar graph of $\textrm{P}$ is topologically equivalent to a bouquet of $k$ circles (for $k$ odd) and of $2 k$ circles for $k$ even. For $r = 1$, the polar graphs are topologically equivalent iff $k$ odd.
(4) Notice that we can preserve some of the qualitative behavior of the examples if we replace $\sin kt$ with any other odd function with period $2 \pi$. In particular, taking the first two interesting terms of the Fourier series of any such function gives pairs
\begin{align}
\tilde \rho &:= 1 + r (\sin k t + a \sin 3 k t ) \\
\tilde{\textrm{P}} &:= R (\sin k t + a \sin 3 k t) .
\end{align}
I'm not sure if @user170231 realized he practically solved the question with his comment. Using his idea one can construct the parametrization
\begin{cases}
\left(\arccos(\cos^2(u))+\arccos(\sin^2(u)),\arccos(\cos^2(u))-\arccos(\sin^2(u))\right) & 0\le u\le \frac{\pi}{2} \\
\left(\arccos(\cos^2(u))-\arccos(\sin^2(u)),\arccos(\cos^2(u))+\arccos(\sin^2(u))\right) & \frac{\pi}{2}\le u\le \pi \\
\left(-\arccos(\cos^2(u))-\arccos(\sin^2(u)),\arccos(\sin^2(u))-\arccos(\cos^2(u))\right) & \pi \le u\le \frac{3 \pi}{2} \\
\left(\arccos(\sin^2(u))-\arccos(\cos^2(u)),-\arccos(\cos^2(u))-\arccos(\sin^2(u))\right) & \frac{3 \pi}{2}\le u\le 2 \pi
\end{cases}
which satisfies the requirement of been $C^2$.
Best Answer
$\def\RR{\mathbb{R}} \def\vr{\mathbf{r}} \def\vv{\mathbf{v}} \def\vnull{\mathbf{0}} \def\ss{\subseteq} \def\To{\rightarrow} \def\p{\pi}$Consider a curve in $\RR^n$ given by $\vr(t) = (x_1(t),\ldots,x_n(t))$ for $t\in I\ss\RR$. Suppose $\vr$ is differentiable and non-self intersecting. (The restriction of non-self intersecting can be relaxed, but introduces technicalities that we wish to avoid.) Let $\vv(t) = \vr'(t)$ and further suppose that $\vv(t)\ne \vnull$ for any $t$. The arclength along this curve will be given by \begin{equation}s=\int_{t_0}^t |\vv(t)|dt,\tag{1}\end{equation} where $t_0\in I$ is some reference value for the parameter $t$ corresponding to an arclength of $s=0$. (Note that $t<t_0$ corresponds to a negative arclength. This sign agrees with the orientation of the curve.) By assumption, $|\vv|>0$. Thus, $s$ is a strictly monotone increasing function of $t$ and so will have an inverse function, $t=t(s)$. Thus, $\vr(s)\equiv\vr(t(s))$, where $s\in I'\ss\RR$, will be this same curve parametrized by arclength. The interval $I'$ can be determined from (1).
After this procedure has been done, it should be clear that two parametrizations of the same curve, $\vr_1(s)$ and $\vr_2(s)$, can differ in their representation only through the following transformation: $s\To ks+s_0$, where $k=\pm 1$. That is, the curves can only differ by a shift in $s$ or by having opposite orientations. (It is also possible that after this transformation further transformations on $I'$ may be allowed and necessary, depending on the periodicity of the components of $\vr(s)$.)
A quick way that one can distinguish different curves is by examining the intervals $I'$. For example, if the intervals are finite subsets of $\RR$ and are different in length, then the curves are not the same. (They do not have the same length.)
Examples:
Let \begin{align*} \vr_1(t) &= (\sin t,\cos t), & t\in[0,2\p) \\ \vr_2(t) &= (\cos 2t,\sin 2t), & t\in[0,\p) \\ \vr_3(t) &= \left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right), & t\in\RR \end{align*} (The last example is @Doug M's from the comments.) It is a straightforward exercise to show that \begin{align*} \vr_1(s) &= (\sin s,\cos s), & s\in[0,2\p) \\ \vr_2(s) &= (\cos s,\sin s), & s\in[0,2\p) \\ \vr_3(s) &= \left(\cos s,\sin s\right), & s\in(-\p,\p) \end{align*} Here we have let $t_0=0$ for each parametrized curve for simplicity. Note that $\vr_1(-s+\p/2) = \vr_2(s)$. Under this transformation, $I_1=[0,2\p)\To[-3\p/2,\pi/2)$. Using periodicity, $I_1\To[0,2\p)$. Thus, the first and second parametrized curves correspond to the same graph. Using periodicity we find $I_3=(-\p,\p)\To[0,2\p)\setminus\{\p\}$. Thus, the third parametrized curve corresponds to a different graph.
As an exercise, using this procedure one should be able to determine that the following curves correspond to the same graph. \begin{align*} &(\sin t,\cos t), & t\in[0,\p] \\ &(\cos 2t,\sin 2t), & t\in[-\p/4,\p/4] \\ &\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right), & t\in[-1,1] \end{align*}