As We are moving in a circle in uniform velocity, so the centripetal force acting on us should be $$ F_{net}= \frac {mv^2}{R} =\frac {4\pi^2mR}{T^2}. $$ There are only two forces acting on us. The normal force and the gravitational force. So
$$ mg-F_N = \frac {4\pi^2mR}{T^2}.$$
Does that mean our actual weight is not mg but $ F_N = mg - \frac {4\pi^2mR}{T^2} $?
That depends on how you define the phrase "actual weight."
If actual weight means the gravitational force of Earth on an object, then $mg$ is correct.
If actual weight means what a scale measures, then your formula that takes into account Earth's rotation is correct. In fact, this would mean actual weight varies with latitude because $R$ gets smaller as you get closer to the poles.
Careful definitions of what is being measured is important for any scientific experiment. For another example, there are two lengths of time that are called a day: a synodic day day (the length of time it takes the sun to return to the same position in the sky) and a sidereal day (the time it takes distant stars to return to the same positon in the sky). Because Earth moves around the Sun as it rotates, these two days are slightly different.
You're mixing two different description of the problem.
The "actual active" force is only the gravitational attraction, $\mathbf{F} = - G \frac{M_E m}{R^2} \mathbf{\hat{r}}$. The reaction of your feet on the ground is the other force acting on you, $\mathbf{N}$.
Assuming the center of the Earth on a inertial reference frame, and constant angular velocity $\boldsymbol{\Omega} = \Omega_E \mathbf{\hat{z}}$, the acceleration of the points on the Earth surface reads,
$\mathbf{a} = \boldsymbol{\Omega} \times \boldsymbol{\Omega} \times \mathbf{r} = - \Omega^2_E R \cos \gamma \, \mathbf{\hat{c}} = -\Omega^2 R \cos^2 \gamma \mathbf{r} + \Omega^2 R \cos \gamma \sin \gamma \mathbf{\hat{n}}$ ,
being $\mathbf{\hat{c}}$ the unit normal vector pointing towards the axis of rotation of the Earth, and $\mathbf{\hat{n}}$ the unit normal vector tangent to the Earth surface and pointing towards North, and $\gamma$ the latitude.
Putting everything together,
$r: \qquad - m \Omega^2 R \cos^2 \gamma = - G \dfrac{M m}{R^2} + N_r$
$n: \qquad - m \Omega^2 R \cos \gamma \sin \gamma = N_n$
So, the gravitational force between you and the Earth is $- G \dfrac{M m}{R^2}$, while the measurement you take as the normal reaction is
$N_r = m \left[ G \dfrac{M}{R^2} - \Omega^2 R \cos^2 \gamma \right] = m \left[ g - \Omega^2 R \cos^2 \gamma \right]$.
Introducing the actual value of Earth angular velocity and radius,
$\Omega = \dfrac{2 \pi}{24 \cdot 60 \cdot 60 s} = 7.3 \cdot 10^{-5} s^{-1}$, $\qquad R = 6.37 \cdot 10^6 m$,
at the equator $\gamma = 0$, you get $\Omega^2 R = 3.36 \cdot 10^{-2} \frac{m}{s^2}$, approximately $0.35 \%$ of a reasonable value of $g = 9.8 \frac{m}{s^2}$.
