Let $\Omega$ is a $C^2$-smooth bounded domain in $R^2$ (not assume to be simple connected) then $f\in W^{1,2}(\Omega) $ can be represented as:
$$f=\nabla^{\perp}b+\nabla\varphi$$
With $b,\varphi\in W^{2,2}(\Omega)$, where$(x,y)^{\perp}=(-y,x)$.
How to prove this? I only know the decomposition in $L^p(\Omega)$ in which $f=u+\nabla p$ and $\text{div} v=0$.
The gradients of $W^{2,2}$ functions form a closed subspace $G$ of the space of $W^{1,2}$ vector fields. Let $\nabla \varphi$ be the orthogonal projection of $f$ onto $G$. The field $f-\nabla \varphi$, being orthogonal to $G$, is divergence-free (a consequence of integration by parts). So, after rotating it by $\pi/2$ we get a curl-free field $F$ from it.
This is where the lack of simple-connectedness becomes annoying: not every curl-free field is the gradient of a function. But this can be fixed by adding to $F$ a harmonic field (one that is both curl-free and divergence-free). Indeed, there are finitely many boundary components of $\Omega$, and the condition for $F$ having a potential amounts to having zero integral over every loop around a component. Add to $F$ a suitable multiple of the field $\nabla^\perp \log|z-z_0| $ (with $z_0$ in a component of complement) to make this integral zero. To balance this off, subtract this multiple of $\log|z-z_0|$ from $\varphi$.