With a bit of trigonometry, we can prove that $a_{\perp}$, the magnitude of the acceleration component perpendicular to the line of sight, is equal to $a \cos(\theta)$. That is why if point D is near the y-axis, the electric field vector there is very small, or zero.

A very good question would be why the line of sight is drawn from point D to the center of the coordinate system, and not from D to the moving charge. This is because of Feynman's approximation in the formula: $$E(t) = -\frac{q}{4\pi\epsilon_0 c^2 r} a_{\perp} \left(t - \frac{r}{c}\right)$$ (Note: I changed $a_x$ to $a_{\perp}$ here because only the projection of the acceleration perpendicular to the line of sight is important). This formula requires the delay time $\Delta t = \frac{r}{c}$ to be constant. Since the speed of light $c$ is constant, the distance $r$ must also be treated as a constant that does not depend on time $t$. This approximation is perfectly valid as long as the charge moves only a microscopically small distance compared to $r$, which is the case here.

The opacity of the electric field vectors is dynamically scaled based on their magnitude. This is a visual effect used to make the propagating waves clearly visible across the canvas.