Now I have understood how to draw the classical unital, $U_q$ at least for $q=3$. It is known that this unital is isomorphic to the subset of the projective plane $PG(2,9)$, defined by the equation $X^4+Y^4+Z^4=0$ in homogeneous coordinates. The projective plane $PG(2,9)$ is just the affine plane $\mathbb F_9\times\mathbb F_9$ with the attached projective line $\mathbb F_9\cup\{\infty\}$. The affine plane and the projective lines are over the $9$-element field $$\mathbb F_9=\{-1-i,-1,-1+i,-i,0,i,1-i,1,1+i\}$$where $i\in\mathbb F_9$ is an element such that $i^2=-1$. The addition in $\mathbb F_9$ is by modulo $3$. The list of the elements of the field $\mathbb F_9$ determines the linear order on $\mathbb F_9$ that will be used in our drawing later.
In the field $\mathbb F_9$, the elements $1$ and $-1$ have four roots of 4-th order: $$\sqrt[4]{1}=\{-1,-i,i,1\}\quad\mbox{and}\quad\sqrt[4]{-1}=\{-1-i,-1+i,1-i,1+i\}.$$
Then the $28$-element classical unital $U_3$ can be identified with the subset $$U_3=(\{0\}\times\sqrt[4]{-1})\cup(\sqrt[4]{-1}\times\{0\})\cup(\sqrt[4]{1}\times\sqrt[4]{1})\cup \sqrt[4]{-1}$$of the projective plane $PG(2,9)=(\mathbb F_9\times\mathbb F_9)\cup\mathbb F_9\cup\{\infty\}$.
In this model, we can draw lines and see whatever we would like to see. For example, that the unital $U_3$ satisfies the O'Nan condition:
$$\forall o,x,y\in U_3\;\forall p\in\overline{xy}\setminus(\overline{ox}\cup\overline{oy})\;\forall u\in \overline{oy}\setminus\{o,y\}\;(\overline{up}\cap\overline{ox}=\varnothing).$$
At the following drawing, this is illustrated on the example of the points $\color{blue}{o=(-1,-1)}$, $\color{blue}{x=(1,-1)}$, $y=(-1,1)$ and $p=(i,-i)\in\overline{xy}$. For the points $\color{magenta}{u=(-1,-i)}$ and $\color{red}{v=(-1,i)}$ on the line $\overline{oy}$ we can see that the lines $\color{magenta}{\overline{up}}$ and $\color{red}{\overline{vp}}$ are indeed disjoint with the line $\color{blue}{\overline{ox}}$.
