Consider polar coordinates, with line element \begin{equation} ds^2 = dr^2 + r^2\,d\phi^2 \end{equation} and (ordered) orthonormal basis $\{dr,r\,d\phi\}$. The orientation is \begin{equation} \omega = dr\wedge r\,d\phi \end{equation} and the computation is nearly identical to the previous one. We have \begin{equation} dr \wedge {*}dr = g(dr,dr)\, dr \wedge r\,d\phi = dr \wedge r\,d\phi \end{equation} from which it follows that \begin{equation} {*}dr = r\,d\phi \end{equation} Similarly, from \begin{equation} r\,d\phi \wedge {*}r\,d\phi = g(r\,d\phi,r\,d\phi)\, dr \wedge r\,d\phi = dr\wedge r\,d\phi \end{equation} we see that \begin{equation} {*}r\,d\phi = -dr \end{equation}

What about $*d\phi$? The Hodge dual is linear, so \begin{equation} *d\phi = *\left(\frac{r}{r}\,d\phi\right) = \frac{1}{r} {*}(r\,d\phi) = -\frac{1}{r} dr \end{equation} Be careful when working with differential forms that are not normalized; it is often easiest to first normalize them, compute the dual of the normalized form, then undo the normalization using linearity.