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What is the formula to compute the signed angle between two vectors $u, vinmathbb{R}^2$ where positive angle is equivalent to a counter-clockwise rotation on the plane

In other words I would like to create a function $angle(u, v)$ where

$$angle(u, v) = -angle(v, u)$$
$$angle(u, v) = -angle(u, -v)$$
$$angle(u, v) = angle(-u, -v)$$
$$angle(u, v) = pi + angle(-u, v)$$

and I want that $angle(e_1, e_2)=frac{pi}{2}$

Just a small observation first, you will have some minor inconsistencies when the angle is either $pi$ or $0$.

To calculate the angle, you can use the $sin$ and $cos$ function, at the same time. Using both will allow you to uniquely define the angle between $-pi$ and $pi$. The cosine of the angle is given by the scalar product:$$ucdot v=|u||v|costheta$$In $mathbb R^2$ you can write the scalar product as $ucdot v=u_xv_x+u_yv_y$.

You get the sine of the angle using cross product $$utimes v=|u||v|sintheta$$
In $mathbb R^2$ you can write the cross product as $utimes v=u_xv_y-u_yv_x$.

Note that the ratio of the cross product and scalar product is the tangent of the angle. But the range of the $arctan$ function is from $-pi/2$ to $pi/2$. Note however that there is a function $mathrm{arctan2}$ or $mathrm{atan2}$ in some programming languages that take the sine and cosine and give you the correct angle. See for example the definition on wikipedia
Then $$angle(u,v)=mathrm{arctan2}(utimes v,ucdot v)$$