Why does direction of vector formula produce same signed angle, even when exchanging the order of points?
For points $P=(x_1,y_1)$, $Q=(x_2,y_2)$
Angle of direction vector is
$\arctan(\theta)=\frac{y_2-y_1}{x_2-x_1}$
If we assume for example $P=(2,3)$, $Q=(5,8)$.
Then both
atan((8-3)/(5-3))*180/pi
and
atan((3-8)/(3-5))*180/pi
= 68.19859 deg
Even though visually they should be different depending on whether one goes from $P$ to $Q$ or from $Q$ to $P$?
Do I need another formula?
In your example, \begin{align*} \frac{y_2 - y_1}{x_2 - x_1} &= {8-3}{5-3} = \frac{5}{2} \\ \frac{y_1 - y_2}{x_1 - x_2} &= {3-8}{3-5} = \frac{-5}{-2} = \frac{5}{2} \\ \end{align*} So it's not really the arctan that's obscuring the order of the points, it's the slope formula.
In the general formula: $$ \tan \theta = m = \frac{y_2-y_1}{x_2-x_1} $$ $\theta$ is the angle measured from the positive $x$-axis to the line with slope $y=mx$. Since $m$ doesn't depend on the order of the points, $\theta$ won't either, not from this formula.
If you want an orientation-dependent vector angle formula in $\mathbb{R}^2$, you can use the cross product. If $\vec u = \left<x_1,y_1\right>$ and $\vec v = \left<x_2,y_2\right>$, then $\vec u \times \vec v = \left<0,0,x_1y_2 - x_2 y_1\right>$ will have a positive $z$-component if $\vec v$ is counterclockwise from $\vec u$, and negative if clockwise. Indeed, $$ \sin\theta = \frac{x_1y_2 - x_2y_1}{\sqrt{x_1^2 + y_1^2} \sqrt{x_2^2 + y_2^2}} $$