How to derive the Equilibrium value $\hat F$ equation
Sign problem

When you go from 

$F_t(1-(1-2m+m^2)-\frac{(1-2m+m^2)}{2N})$

to

$F_t(-2m+m^2-\frac{(1-2m+m^2)}{2N})$

You make a mistake in the sign. It should be

$F_t(2m-m^2-\frac{(1-2m+m^2)}{2N})$

Approximations

I will assume this error of sign propagate to inverse your end result. You hence should get

$F_t= \frac{1}{4Nm+1-2m} -\frac{2m}{(-4Nm)-1+2m}$ 

$\frac{2m}{4Nm+1+2m}$ is negligible compared to $\frac{1}{4Nm+1+2m}$, so you can indeed remove it. You get

$F_t= \frac{1}{4Nm+1-2m}$ 

Also, $2m$ is negligible compared to $4Nm$, so again you can remove it. You end up with

$F_t= \frac{1}{4Nm+1}$ 

Tadaa!