Net force when deriving relation between torque and angular acceleration

You are confusing a 2D problem with a 1D problem. In 2D, force and motion also have direction (other than 1 or -1). The two forces you are trying to sum are orthogonal to each other and should not appear in each other's force equilibrium.

To put it in coordinates, let's choose a $(r,theta)$ coordinate system. In that system $$vec{F}=F_rhat{r}+F_{theta}hat{theta}.$$

To calculate the torque about the origin that this force exerts on a particle located at position $vec{r}$ we calculate a cross product $$vec{tau}=vec{r}timesvec{F}.$$

In case you don't know about cross products, one of the results is that any component of $vec{F}$ that is parallel to $vec{r}$ vanishes, and because $vec{r}$ points in the $hat{r}$ direction, the $F_r$ component doesn't contribute to the torque.$$vec{tau}=rF_{theta}hat{k}$$

For a more intuitive idea, consider this. If you have a particle at rest some distance from the origin and you exert a force on it directly toward the origin, there is no angular acceleration around the origin. By definition, angular accelerations must be tied to a torque, and vice versa, so zero angular acceleration implies zero torque. Radial forces don't cause torques about the central point.

$1^{st}$ method

The torque is only generated by the perpendicular component of force acting along the radius vector (as can be seen from this formula $ mathbf tau = mathbf r times mathbf F $)

---

$2^{nd}$ method

You may know that $v neq omega r$ but $mathbf v = mathbf omega times mathbf r $. Therefore $mathbf a = mathbf alpha times mathbf r $. Now you might notice that since the component of $mathbf a$ which is parallel to $ mathbf r$ can never be generated via a cross product therefore the component of acceleration parallel to the radius isn't accounted for.