Let us consider firstly the set of parametric plots in which the coordinates $x,y$ are given by $ [\sin(t), \cos(kt)]$. I will restrict my answer to the cases where $k$ is an integer.
In this set of parametric plots, the function crosses the $x$-axis for $2k$ different values of $t$, which are those satisfying $t=\pi(j-1/2)/k$, where $j=1,2,3...2k$. However, because of the symmetry of these $t$ values with respect to the $y$-axis, these zeros can be grouped in pairs with the same $x$-value, i.e. corresponding to the same point on the $x$-axis. As a result, the function crosses the $x$-axis in $k$ points when $k$ is even, and $k+1$ points when $k$ is odd.
For example, when $k=4$, the function crosses the $x$-axis for eight values of $t$, namely $\frac{\pi}{8}, \frac{3}{8}\pi, \frac{5}{8}\pi... \frac{15}{8}\pi$. However, these eight values correspond to only four points on the $x$-axis, since the $x$ value is equal for the pair $t=\frac{\pi}{8}$ and $t=\frac{7}{8}\pi$, for the pair $t=\frac{3}{8}\pi$ and $t=\frac{5}{8}\pi$, and so on. Accordingly, the parametric plot $ [\sin(t), \cos(4t)]$ is as follows:
On the other hand, for example, when $k=3$, the function crosses the $x$-axis for six values of $t$, namely $\frac{\pi}{6}, \frac{3}{6}\pi, \frac{5}{6}\pi... \frac{11}{6}\pi$. These six values correspond to only four points on the $x$-axis, since the $x$ value is equal for the pair $t=\frac{\pi}{6}$ and $t=\frac{5}{6}\pi$, and for the pair $t=\frac{7}{6}\pi$ and $t=\frac{11}{6}\pi$; the other two points are given by $t=\frac{3}{6}\pi$ and $t=\frac{9}{6}\pi$. Accordingly, the parametric plot $ [\sin(t), \cos(3t)]$ is as follows:
Note then that, when $k$ is even, all $2k$ zeros of the function can be grouped in pairs, leading to $k$ points on the $x$-axis; conversely, when $k$ is odd, all $2k$ zeros of the function can be grouped in pairs, except two that are "unpaired" (those corresponding to $t=\frac{\pi}{2}$ and $t=\frac{3}{2}\pi$). This leads to $k+1$ points on the $x$-axis. Also note other two things: 1) for $k$ odd, the two "unpaired" zeros given by $t=\frac{\pi}{2}$ and $t=\frac{3}{2}\pi$ correspond to the most lateral points (left and right) of the function; 2) for $k=1$, the function crosses the $x$-axis only in these two unpaired zeros, thus without crossing itself (in fact for $k=1$ it reduces to a circle).
Now it is also not difficult to show that, considering the behaviour of the function in these points laying on the $x$-axis, its derivative has a unique value when $k$ is even - which means that the function does not cross itself. On the other hand, when $k$ is odd, the derivative assumes two different values (positive and negative) in each of these points laying on the $x$-axis except for the two "unpaired", most lateral ones. This means that the function crosses itself in $k-1$ points on the $x$-axis. For example, we expect that for $k=14$ the parametric plot $ [\sin(t), \cos(kt)]$ crosses the $x$-axis in $14$ points and does not cross itself. This is demonstrated by the following figure showing $ [\sin(t), \cos(14t)]$:
On the other hand, for example we expect that for $k=9$ the parametric plot $ [\sin(t), \cos(kt)]$ crosses the $x$-axis in $9+1=10$ points, and that it crosses itself in $9-1=8$ of these points. This is demonstrated by the following plot of $ [\sin(t), \cos(9t)]$, showing the $8$ points where the function crosses itself and the remaining $2$ most lateral points ("unpaired" zeros) where the function crosses the $x$-axis:
As a result, within the set of parametric plots $x,y$ given by $ [\sin(t), \cos(kt)]$, there is no possibility to find a case that yields a unique crossing point dividing the function in two sides (left and right). For $k$ even, the function does not cross itself; among the odd $k$ values, for $k=1$ the function does not cross itself; for $k=3$, the function crosses the x-axis in four points (two of which are the lateral, "unpaired" ones), and therefore crosses itself in two points on the $x$-axis; lastly, as already shown, for higher odd values of $k$, the function crosses itself on the $x$-axis even more times.
All these considerations can be repeated for the set of parametric plots in which the coordinates $x,y$ are given by $ [\sin(t), \sin(kt)]$, with the only difference that the behaviour of the function for $k$ odd or even is inverted. More precisely, in this case the function crosses the $x$-axis $k$ times and never crosses itself when $k$ is odd; conversely, the function crosses the $x$-axis in $k+1$ points and crosses itself in $k-1$ of these points when $k$ is even.
As a result, within the set of parametric plots $x,y$ given by $ [\sin(t), \sin(kt)]$, the only case in which there is a unique crossing point dividing the function in two sides (left and right) occurs for $k=2$. In fact, in this case, the function crosses the $x$-axis in three points (two of which are the"unpaired" lateral ones) and crosses itself in a single central point, yielding a "twisted" ball according to the definition given in the OP. Accordingly, the plot of $ [\sin(t), \sin(2t)]$ is as follows:
