In Cartesian coordinates, given a one-form $\alpha$, the components of $d\alpha$ give the curl of the vector field $V=\alpha^\sharp$. That is,
$$\textrm{curl}\left(\alpha^\sharp\right)=\left(\star d\alpha\right)^\sharp$$
I'm trying to use this to get the form of curl in cylindrical coordinates, but I seem to be getting the wrong answer. Let
$$\alpha =\alpha_r dr+\alpha_\phi d\phi+\alpha_z dz$$
where $x=r\cos\phi$, $y=r\sin\phi$, $z=z$.
$$d\alpha= \left(\frac{\partial\alpha_\phi}{\partial r}-\frac{\partial\alpha_r}{\partial \phi}\right) dr\land d\phi+\left(\frac{\partial\alpha_z}{\partial \phi}-\frac{\partial\alpha_\phi}{\partial z}\right) d\phi\land dz+\left(\frac{\partial\alpha_z}{\partial r}-\frac{\partial\alpha_r}{\partial z}\right) dr\land dz$$ then $$\star d\alpha= r\left(\frac{\partial\alpha_z}{\partial \phi}-\frac{\partial\alpha_\phi}{\partial z}\right) dr-r\left(\frac{\partial\alpha_z}{\partial r}-\frac{\partial\alpha_r}{\partial z}\right) d\phi+r\left(\frac{\partial\alpha_\phi}{\partial r}-\frac{\partial\alpha_r}{\partial \phi}\right) dz$$
Finally, $$\left(\star d\alpha\right)^\sharp=r\left(\frac{\partial\alpha_z}{\partial \phi}-\frac{\partial\alpha_\phi}{\partial z}\right) \left(\frac{\partial }{\partial r}\right)-r^{-1}\left(\frac{\partial\alpha_z}{\partial r}-\frac{\partial\alpha_r}{\partial z}\right) \left(\frac{\partial }{\partial \phi}\right)+r\left(\frac{\partial\alpha_\phi}{\partial r}-\frac{\partial\alpha_r}{\partial \phi}\right) \left(\frac{\partial }{\partial z}\right)$$
So, if $\alpha^\sharp=V_r \left(\frac{\partial }{\partial r}\right)+V_\phi \left(\frac{\partial }{\partial \phi}\right)+ V_z \left(\frac{\partial }{\partial z}\right)$ then
$$\textrm{curl}\left(V\right)=r\left(\frac{\partial V_z}{\partial \phi}-\frac{\partial \left(r^2 V_\phi\right)}{\partial z}\right) \left(\frac{\partial }{\partial r}\right)-r^{-1}\left(\frac{\partial V_z}{\partial r}-\frac{\partial V_r}{\partial z}\right) \left(\frac{\partial }{\partial \phi}\right)+r\left(\frac{\partial \left(r^2 V_\phi\right)}{\partial r}-\frac{\partial V_r}{\partial \phi}\right) \left(\frac{\partial }{\partial z}\right)$$
This however, doesn't match wikipedia. What went wrong?
You need to write your $1$-form in terms of unit length basis covectors, namely $dr$, $r\,d\phi$, and $dz$, and then figure out the $\star$ ... It looks to me like you said $\star(d\phi\wedge dz) = r\,dr$. In fact, $\star(r\,d\phi\wedge dz) = dr$, so $\star(d\phi\wedge dz) =\dfrac1r\,dr$. Similarly with $\star(dr\wedge d\phi)$. I haven't checked the rest. (P.S. You should work with an orthonormal basis for the tangent vectors, too.)