Take $v$ a smooth incompressible vector field and call $w$ its vorticity.

If $w\in L^1(\mathbb{R}^2)$, the number \begin{align*} \alpha := \int_{\mathbb{R}^2} w \,dx, \end{align*} is well-defined.

First fix $\psi\in\mathscr{D}(\mathbb{R}_+^*)$ (i.e. smooth and compactly supported) such as \begin{align*} \int_0^\infty \psi(r)\,r\,dr = \alpha. \end{align*}

Since $\psi$ vanishes totally in $0$ (in particular $\psi'(0)=0$ which is what we need here), you have the existence of a smooth function $\overline{w}$ such as $\psi'(x)=x \overline{w}(x)$.

Define the radially symmetric vector field $b(x):=\psi(\|x\|)$, so that \begin{align*} \int_{\mathbb{R}^2}b\, dx = \alpha. \end{align*}

Remark that you have also \begin{align*} b(x) = \int_0^{\|x\|} \overline{w}(s)\,s \,ds. \end{align*} Now, by construction the vector field $u:=v-K_2* b$ satisfies \begin{align*} \text{div}\,u &= 0,\\ \int_{\mathbb{R}^2} \text{curl}\,u &= 0, \end{align*} so that thanks to the Taylor expansion you mentionned, one has $u\in L^2(\mathbb{R}^2)$.

Take $v$ a smooth incompressible vector field and call $w$ its vorticity.

If $w\in L^1(\mathbb{R}^2)$, the number \begin{align*} \alpha := \int_{\mathbb{R}^2} w \,dx, \end{align*} is well-defined. Looking at Example 2.1. on p.47 (so-called "radial eddies"), we now that for any function $\overline{w}:\mathbb{R}_+\rightarrow\mathbb{R}$, the vector field $b$ that you defined is divergence free, with curl $x\mapsto \overline{w}(|x|)$, so you just have to choose $\overline{\omega}$ properly to insure that \begin{align*} \int_{\mathbb{R}^2} \overline{w}(|x|)\,dx = 2\pi \int_0^\infty r \,\overline{w}(r)dr = \alpha, \end{align*} and you're done since $u:=v-b$ is then divergence-free vector field having a vorticity with $0$ mean on $\mathbb{R}^2$ : the Taylor expansion you mentionned allows to conclude that $u\in L^2(\mathbb{R}^2)$.