This is not intended as a full answer, but as a way to try to decide if it is feasible to be able to parameterize a plane curve in an easy way.
The two examples you give, namely the folium and the bicorn, are examples of "irreducible curves of geometric genus zero". For the purpose of this discussion you can take "irreducible curve" to mean the locus of zeros of an irreducible polynomial in two variables. Geometric genus zero is harder to define, though see this question and the various answers, especially the answer of Matt E. In the linked answer, the "genus" that is defined is the "arithmetic genus", whereas we are interested in the "geometric genus".
The two genera are not unrelated, and the discrepancy between them is concentrated at the singular points of the curve. A singular point of a plane curve $C$ given as the vanishing set of a polynomial $f(x,y)$ is a point $p$ in the plane that lies on the curve (i.e. $f(p)=0$) and where both partial derivatives of $f$ vanish (i.e. $df/dx(p)=df/dy(p)=0$).
Thus, while the arithmetic genus of an (irreducible) plane curve $C$ of degree $d$ is simply $p_a:=(d-1)(d-2)/2$, the geometric genus of $C$ is an integer less than or equal to $p_a$. I will not dwell too much on how to compute the geometric genus, for the moment I will simply say that every singular point decreases the arithmetic genus by at least one (counting also the "esoteric" singular points at infinity, as you need to explain the bicorn). There are algorithmic way to decide the drop in genus that every singular point determines and thus the computation of the geometric genus can be taken as "easy".
Over the complex numbers. It is a fact that, whenever the geometric genus of an irreducible curve $C$ is zero then you can find a parameterization of $C$ by rational functions of a single variable $t$, provided that you are thinking about the complex solutions of your curve. This parameterization can be found algorithmically, and while there are various tricks to use, the substitution $y=mx$ is at the core: depending on context it might be called "projection" or "blow up". The other trick is that you can "reembed" your curve by considering the vector space whose coordinates are all the monomials of some given degree $n$ and take the image of the curve under the map that sends a point in the plane in the $N$-tuple of all the evaluations of the monomials you have chosen. Instead of being more explicit, let me give an example: if the curve is the $x$-axis in the plane, and you choose the monomials $1,x,y,x^2,xy,y^2$ to reembed, then we image of the $x$-axis (i.e. the locus where $y=0$) is the set of points with coordinates $(1,x,0,x^2,0,0)$ in a 6-dimensional space. Forgetting the useless coordinates (this is an instance of projection) gives us a plane curve parameterized by $(x,x^2)$, namely the parabola $x_1^2=x_2$. In this example we started with a parameterization and ended with a parameterization. Being more refined you can also start with an implicit equation and end with (lots of) implicit equations in more variables.
Once you have projections and reembeddings you can play them against one another: embed your curve in large enough space. Project away from singular points until you get back to the plane; reembed and reproject. There is a way of making sure that "singular points get resolved" by doing this, so that eventually you get to a curve having no singular points. Once you are there, further projections will do the trick of getting you to a line.
Over the real numbers. Of course things are somewhat trickier. Certainly two possibilities are the folium, where the rational parameterization you had over the complex numbers "just works" over the reals, but there is the slightly more elaborate one of the bicorn, where you have a rational parameterization in terms of trigonometric functions. Again, by modification of the tricks discussed above, you can get every irreducible curve of genus zero to become a plane conic. Once in this form, you will find a rational parameterization if and only if you can find a point, and deciding if there is a point reduces to the computation of whether a matrix constructed out of the coefficients of the equation has eigenvalues of different signs or not. The distinction between rational parameterization in terms of a variable or in terms of trigonometric functions has to do with personal taste.
So far, this has dealt with curves of genus zero and I will not say much about higher genus. If the genus is one, there is, as you mentioned, the possibility of using the Weierstrass $\wp$. Here things become more complicated: the function used to parameterize everything is less familiar, you will probably need some transcendental numbers to get this parameterization to work, and the topological description of the space of solutions gets more in the way. In the genus zero case, the only thing that mattered was: is it empty, is it not empty. In the genus one case there is "one more" possibility. Of course, higher genera become harder and harder, and I will stop here, given that this is already way too much for this!
Best Answer
The comment by @Blue is correct. I've had experience with these parallel curves and plotted them up for the bean curve in the first figure below. The first algorithm in the Wikipedia article was used, but be advised that the algorithm breaks down if the $|d|$ is too large or too small. Here is what I did: I first centered the bean curve at its circumcenter. Then I normalized it to fit in a unit circle. For the parallel curves I took $d\in [-.3,.3]$ in $0.1$ increments. In the complex plane, let's call the bean curve $z(t)$ and the parallel curves $z_d(t)$. The we have
$$ z(t)=(\cos^3t+\sin^3t)e^{it}\\ z_d(t)=z(t)-d\frac{i\dot z}{|\dot z|} $$
where $\dot z=\dot x+i\dot y$. The Cartesian equations are given in the referenced Wiki article.
It's interesting to note the circumcircles of all the parallel curves are concentric (they share the same center) but that the circumcenters of al the beans are in different relative positions. This is demonstrated in the second figure below.