Start with a finite field of order $4$:
+ & 0 & 1 & a & a+1 \\\hline
0 & 0 & 1 & a & a+1 \\
1 & 1 & 0 & a+1 & a \\
a & a & a+1 & 0 & 1 \\
a+1 & a+1 & a & 1 & 0
\times & 0 & 1 & a & a+1 \\\hline
0 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & a & a+1 \\
a & 0 & a & a+1 & 1 \\
a+1 & 0 & a+1 & 1 & a
Now you can draw the points as a $4\times4$ grid using these elements as $x$ and $y$ coordinate. That's the $n^2$ points you'd expect from an affine plane of order $n$. You can also draw lines of constant $x$ coordinate (i.e. vertical lines), and non-vertical lines with equations $y=tx+u$ where $t$ and $u$ are again elements from your field. This is $n+n^2$ lines, as expected.
It doesn't really matter how you draw the lines, as long as the points on each line come from the same equation. If the lines you draw intersect in points besides the 16 points in the plane, that simply doesn't count as an intersection. So one possible illustration would be this:
Here I've drawn lines of the same “slope” (i.e. same parameter $t$) in the same color. Cyan is $t=0$, navy blue is $t=1$, magenta is $t=a$ and dark green is $t=a+1$. The bright green lines are vertical, $t=\infty$ if you will.