Harmonic functions and polar differential forms Given a harmonic function $u$, its differential and conjugate differential are
$$du = \frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy,\qquad ^{*}du = -\frac{\partial u}{\partial y}dx + \frac{\partial u}{\partial x}dy.$$ We also know that Laplace's equation takes the form
$$r\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right) + \frac{\partial^{2}u}{\partial\vartheta^{2}} = 0$$
when written in polar coordinates.


How do we obtain the polar form of the conjugate differential
    $$^{*}du = r\left(\frac{\partial u}{\partial r}\right)d\vartheta?$$


EDIT:
This is taken from Ahlfors, and it seems that he uses the $^{*}$ to indicate that $^{∗}du$ satisfies
$$f\,dz=du+i\,^{∗}du,$$
where $f=\frac{\partial u}{\partial x} + i\,\frac{\partial u}{\partial y}$.  Also, he states that the form of $^{∗}du$ given holds for a circle $\lvert z\rvert=r$, if that makes a difference.


Also, he states that the form of $∗du$ given holds for a circle $|z|=r$, if that makes a difference.

Oh yes, it does make a difference. This is why context matters. A differential form is a device that eats vectors and produces numbers. For example, $du$ is the form that takes a vector $\vec a$  and returns the directional derivative of $u$ in the direction $\vec a$. And $*du$ is the form that takes a vector $\vec a$, rotates it clockwise by $90$ degrees,  and returns the  derivative of $u$ in that direction. Indeed, 
$$
 -\frac{\partial u}{\partial y}a_1 + \frac{\partial u}{\partial x}a_2 = 
\nabla u \cdot \langle a_2,-a_1\rangle
$$
When we integrate a differential form along a curve, we feed the tangent vector of the curve into the form. Consider a tangent vector to $|z|=r$, traveled counterclockwise. Rotating  it by $90$ degrees clockwise turns the vector into outward normal.  So, $*du$ returns the normal derivative of $u$, namely $\dfrac{\partial u}{\partial r}$. The factor $r\,d\theta$ is the length element of the curve.

Old answer, before the question was edited
The formulas for $du$ and $*du$ have nothing to do with $u$ being harmonic. We can apply the Hodge star operator to any $k$-form in $n$ dimensions, getting an $(n-k)$-form. Here we have the special case of $k=1$ and $n=2$. As long as  we deal with first derivatives, harmonicity does not come into play. 
To express $du$ and $*du$ in polar coordinates,  begin with
$$du = \frac{\partial u}{\partial r} dr +  \frac{\partial u}{\partial \theta}d\theta$$  and then  form $*du$  using the defining property of Hodge star, $\alpha\wedge *\beta = \langle \alpha,   \beta\rangle \, \omega$. Here $\omega$ is the volume form, which in polar coordinates is $r\,dr\wedge d\theta$. So,
$$*\left( \frac{\partial u}{\partial r} dr \right) = r \frac{\partial u}{\partial r} d\theta$$
and 
$$*\left( \frac{\partial u}{\partial \theta} d\theta \right) = -r \frac{\partial u}{\partial \theta} dr$$
The end result is 
$$ *du = -r \frac{\partial u}{\partial \theta} dr + r \frac{\partial u}{\partial r} d\theta
$$ which has an extra term compared to yours. 
