Part I.
I am assuming that the polar coordinates are related to the Cartesian coordinates
by
$$
x=r\sin\theta\cos\phi\,,\quad y=r\sin\theta\sin\phi\,,\quad z=r\cos\theta
$$
so that
$$
f(r,\theta,\phi)=\phi=\arccos\Big(\frac{x}{\sqrt{x^2+y^2}}\Big)\,.
$$
From this answer we know that
$$
df=d\phi=\frac{x\,dy-y\,dx}{x^2+y^2}\,.
$$
The Hodge dual of this is
$$
\star df=\frac{x\,dz\wedge dx-y\,dy\wedge dz}{x^2+y^2}\,.
$$
The right hand side of this is $d\omega$ which can be written as
$$
d\omega=dz\wedge\frac{x\,dx+y\,dy}{x^2+y^2}\,.
$$
This shows that
$$
\omega=-\frac{\log(x^2+y^2)}{2}\,dz\,.
$$
The function $-\frac{\log(x^2+y^2)}{2}$ is not the product of two delta functions $\delta(x)\delta(y)$ but it has a pole at the origin and its integral over the unit disc is
$$
-\pi\int_0^1\log (r^2)\,r\,dr=-2\pi\int_0^1(\log r)\,r\,dr
=2\pi\int_0^1 \frac{r^2}{2r}\,dr-2\pi(\log r)r^2\Big|_0^1=\frac{\pi}{2}\,.
$$
Part II.
You want to know if
$$\tag{1}
\epsilon_{ijk}\,[\partial_{j},\partial_{k}]\,\phi=\hat z_i\,\delta(x)\delta(y)
$$
is true. The left hand side of this contains the gradient of $\phi$ which we know from above is
$$
\nabla\phi=\partial_k\,\phi=\frac{1}{x^2+y^2}\begin{pmatrix}-y\\x\\0 \end{pmatrix}\,.
$$
Since $\epsilon_{ijk}\,a_j\,b_k=(\mathbf{a}\times\mathbf{b})_i$ we have
$$
\epsilon_{ijk}\,\partial_j\,\partial_k\,\phi=(\nabla\times(\nabla\phi))_i
$$
which is zero for all $x,y\not=0\,.$ Therefore,
$$
\epsilon_{ijk}\,[\partial_j,\partial_k]\,\phi=0\,\quad\text{ for all }x,y\not=0\,.
$$
Let $S^1$ be the unit circle around the origin in the $xy$-plane. Since
$$
\int_{S^1}d\phi=\int_0^{2\pi}\,d\phi=2\pi
$$
there seems a contradiction to the Stokes theorem because from $dd\phi=0$ it should follow that this integral must be zero. The problem is however
that $\phi$ is undefined in the origin and the Stokes theorem can be
"preserved" when in the distributional sense the rotation of $d\phi$ is
given by those delta functions in (1).
