Of course, if you break $\mathbb{R}^2$ into a polar grid
$\hspace{3.5cm}$
the small slightly curved rectangles have area $r\,\mathrm{d}\theta\,\mathrm{d}r$.
However, it seems that you are interested in looking at
$$
\begin{align}
\mathrm{d}y\,\mathrm{d}x
&=(\sin(\theta)\,\mathrm{d}r+r\cos(\theta)\,\mathrm{d}\theta)(\cos(\theta)\,\mathrm{d}r-r\sin(\theta)\,\mathrm{d}\theta)\\
&=r\,\mathrm{d}\theta\,\mathrm{d}r
\end{align}
$$
and why the $\mathrm{d}r^2$ and $\mathrm{d}\theta^2$ terms disappear and the $\mathrm{d}r\,\mathrm{d}\theta$ and $\mathrm{d}\theta\,\mathrm{d}r$ have different signs.
Let's start with
$$
\begin{align}
\mathrm{d}x&=\cos(\theta)\,\mathrm{d}r-r\sin(\theta)\,\mathrm{d}\theta\\
\mathrm{d}y&=\sin(\theta)\,\mathrm{d}r+r\cos(\theta)\,\mathrm{d}\theta
\end{align}
$$
rewritten as
$$
\begin{bmatrix}\mathrm{d}x\\\mathrm{d}y\end{bmatrix}
=\begin{bmatrix}\cos(\theta)\\\sin(\theta)\end{bmatrix}\mathrm{d}r
+\begin{bmatrix}-r\sin(\theta)\\r\cos(\theta)\end{bmatrix}\mathrm{d}\theta
$$
Therefore, the displacements $\color{green}{\mathrm{d}r}$ and $\color{red}{\mathrm{d}\theta}$ get mapped to $\color{green}{\begin{bmatrix}\cos(\theta)\\\sin(\theta)\end{bmatrix}\mathrm{d}r}$ and $\color{red}{\begin{bmatrix}-r\sin(\theta)\\r\cos(\theta)\end{bmatrix}\mathrm{d}\theta}$ in $\mathbb{R}^2$:
$\hspace{3cm}$
where the area in gray is given by $\color{green}{\begin{bmatrix}\cos(\theta)\\\sin(\theta)\end{bmatrix}\mathrm{d}r}\times\color{red}{\begin{bmatrix}-r\sin(\theta)\\r\cos(\theta)\end{bmatrix}\mathrm{d}\theta}=r\,\mathrm{d}r\,\mathrm{d}\theta$.
The fact that the cross product is involved is the reason that the $\mathrm{d}r^2$ and $\mathrm{d}\theta^2$ terms disappear and the $\mathrm{d}r\,\mathrm{d}\theta$ and $\mathrm{d}\theta\,\mathrm{d}r$ have different signs. This, and its $n$-dimensional analogs, are why we use wedge products and differential forms when changing variables.
The Jacobian map is for transforming vectors expressed in terms of one set of coordinate basis vectors into another coordinate system's basis vectors. Positions like $(x,y)$ and $(r,\theta)$ are not expressed in terms of coordinate basis vectors, so it's inappropriate to use the Jacobian to try to convert between them.
Let $e_1, e_2$ be a pair of basis vectors. We can express positions on the 2d plane as $p = x e_1 + y e_2$.
Now, let $f(p) = p' = r e_1 + \theta e_2$. This looks like a change of coordinates, but it's really not--it's an active deformation of the plane into something where $r, \theta$ are "Cartesian" coordinates. This is just an active transformation, however, and fully equivalent to the passive change of coordinates that you're used to.
$f$ is appropriate to move positions to new positions, but it is not appropriate to move, for example, the tangent vector to a curve from one space to another (that is, to express such a tangent vector in terms of the polar coordinate basis vectors). For this, we need the Jacobian map $J_f$.
Example: let $\ell(t) = e_1 \cos t + e_2 \sin t$ be a curve that draws out the unit circle. It's clear that its derivative is the tangent vector $\dot \ell(t) = -e_1 \sin t + e_2 \cos t$. We can't transform this tangent vector using $f$; we must use $J_f$ instead.
(You'll note here I'm moving from Cartesian coordinates to polar, backwards from what you wanted* but the math is basically the same.)
Here's the Jacobian map:
$$\begin{align*}
J_f(e_1) &= \frac{x e_1}{\sqrt{x^2 + y^2}} - y e_2 \\ J_f (e_2) &= \frac{y e_1}{\sqrt{x^2 + y^2}} + x e_2\end{align*}$$
Along the curve, $x = \cos t$ and $y = \sin t$, so we get
$$\begin{align*} J_f (\dot \ell(t)) &= -(\sin t )(e_1 \cos t - e_2 \sin t) + (\cos t)(e_1 \sin t + e_2 \cos t) \\ &= e_2\end{align*}$$
Remember that $e_2$ is associated with $\theta$--this says that, unsurprisingly, the velocity is entirely in the $\theta$ direction along this curve. We conclude that $\dot \ell(t) = J_f^{-1}(e_2) = e_\theta$.
In conclusion, we started with a tangent vector $\dot \ell(t)$ in our Cartesian coordinate system, and we moved it--using the Jacobian $J_f$--into a deformed plane where $(r,\theta)$ are "Cartesian" coordinates instead. The Jacobian is what moves tangent vectors from one space to another (or between coordinate systems), but positions are different and will always be handled by the full, nonlinear transformation.
One way you can remember this is that the Jacobian is like the derivative of the transformation, and so it's appropriate for moving things involving derivatives, like $\dot \ell(t)$, which is a velocity.
Best Answer
When you do polar plots you are stuck with parametrizations of the limited form $$x(\varphi)=r(\varphi)\cos \varphi, \quad y(\varphi)=r(\varphi)\sin(\varphi).$$ The parametrization that you gave is not of that form. This is apparent already from the observation that the parametrization gives the full trefoil, when $t\in[0,2\pi]$, but the trefoil wraps around the origin twice, so $\varphi$ should range over $[0,4\pi]$.
Let us look at the Wikipedia parametrization of the trefoil on the surface of a torus: $$ x=(2+\cos3t)\cos2t,\quad y=(2+\cos3t)\sin2t,\quad z=\sin 3t. $$ If we ignore that $z$-coordinate for a moment, we see $(x,y)\uparrow\uparrow(\cos 2t,\sin 2t)$, which is a tell-tale sign that here $\varphi=2t$ is the polar angle coordinate. As $t$ ranges over $[0,2\pi]$, we should, indeed, have $\varphi\in[0,4\pi]$. Thus the projection of that trefoil onto the $xy$-plane comes from the polar equation $$ r=2+\cos\frac{3\varphi}2, $$ as suggested by heropup (+1). The plot is not quite what you may have expected:
Here the additive constant $2$ represents the ratio of the radius of the "wire" inside the torus to that of the "tube" around the wire. IMVHO the projection looks a bit cleaner, if we use ratio $4$ and equation $r=4+\cos\frac{3\varphi}2$ instead:
For a better view here is a 3D-image of how the trefoil wraps itself around the torus.
The trefoil is the thin tube on the surface of the doughnut.