I take your question as asking for an explicit formula on a closed $3$-manifold? In this case, the most interesting example I know is the following:

The standard contact form on the $3$-sphere $\mathbb{S}^3$ in $\mathbb{C}^2$ is

$$ \alpha_0 := i\, \sum_{j=1}^2 (z_j\,d\bar z_j - \bar z_j\, dz_j) \;.$$

Choosing the function $f(z_1,z_2) = z_1^2 + z_2^2$, one can verify that $\alpha_0$ is adapted to the open book $(K,\vartheta)$ with binding
$K = \{f=0\}$ and fibration over the circle given by $\vartheta = f/|f|$.
The page of this open book is $T^*\mathbb{S}^1$ and the monodromy is a *positive* Dehn twist.

If we now subtract the term $i\,(f\,d\bar f - \bar f\, df)$ from $\alpha_0$, we obtain
$$ \alpha_- := i\, \sum_{j=1}^2 (z_j\,d\bar z_j - \bar z_j\, dz_j) - i\,(f\,d\bar f - \bar f\, df) \;,$$

which is still a contact form, it is still supported by an open book with binding $K=f^{-1}(0)$ and with identical pages, but the induced orientation of the $3$-sphere is now different, and thus also the coorientation of the pages. The open book is then $(K,\bar\vartheta = \bar f/|f|)$, i.e., it has the same pages, but its monodromy is a *negative* Dehn twist.

You can also find the overtwisted disk explicitly by intersecting the $3$-sphere with a $3$-dimensional subspace of $\mathbb{C}^2$: The binding $K$ of the open book is given by the two circles
$$K = \bigl\{(z,- i z)\in \mathbb{S}^3\bigr\}\sqcup
\bigl\{(z,+ i z)\in \mathbb{S}^3\bigr\}\;.$$
Let $E\subset \mathbb{C}^2$ be the $3$-space that is spanned (over $\mathbb{R}$) by the complex line $(z,-iz)$ and the real line $(t,it)$. The intersection of $E$ with $\mathbb{S}^3$ is clearly a $2$-sphere. Its equator is one of the components of the binding; the north and south poles are $\pm(1,i)$, that is, the poles are the intersection of $\mathbb{S}^2$ with the second binding component.

Parametrize the $2$-sphere by
$$\Phi\colon (\varphi, \vartheta) \mapsto \frac{1}{\sqrt{2}}\, \sin \vartheta \cdot \begin{pmatrix}e^{i\varphi} \\ -ie^{i\varphi}\end{pmatrix} + \frac{1}{\sqrt{2}}\,\cos\vartheta\cdot \begin{pmatrix}1 \\i \end{pmatrix}$$
with $\varphi\in[0,2\pi]$ and $\vartheta\in [0,\pi]$. The pull-back of $\alpha_0$ with this map is:
$$
\Phi^*\alpha_0 = 2 \sin^2 \vartheta\, d\varphi
$$
so that the characteristic foliation for $\alpha_0$ is just that of a $2$-sphere with segments running down from the north pole to the south pole.

We obtain $f\bigl(\Phi(\varphi, \vartheta)\bigr) = 2\, \cos\vartheta\, \sin\vartheta \, e^{i\varphi}$, so that $\Phi^*(f\,d\bar f - \bar f\, df) = -8i\,\cos^2\vartheta\, \sin^2\vartheta\, d\varphi$ and thus
$$\Phi^*\alpha_- = 2\sin^2\vartheta\, \bigl(1 -4\,\cos^2\vartheta \bigr)\, d\varphi \;.$$
This shows that the $2$-sphere contains an overtwisted disk in the northern hemisphere and a second one in the southern hemisphere, because $\Phi^*\alpha_-$ becomes singular along the two circles $\cos\vartheta = \pm 1/2$ (ie $\vartheta = \pi/3$ or $\vartheta = 2\pi/3$).