Surely you know that a "fixed point" is solution that does not change- a constant solution. And for any constant, the derivative is 0. That is, at a fixed point, dR/dt= BR- ARF= 0 and dF/dt= -DF+ ARF= 0.
We want to solve BR- ARF= R(B- AF)= 0 and -DF+ ARF= F(-D+ AR)= 0.
A product is 0 if and only if at least one of its factors is 0. So for the first equation, we must have either R= 0 or B- AF= 0. If R= 0, the second equation becomes F(-D)= 0 so that D= 0. One fixed point is R= 0, F= 0.
If B- AF= 0, that is if F= B/A, the second equation become (B/A)(-D+ AR)= -BD/A+ BR= 0 so R= D/A. Another fixed point is R= D/A, F= B/A.
Any nonlinear functions could be written as a power series we know that, for the variables close to 0, higher powers are smaller than the variables themselves so can be dropped. About R= 0, F= 0, that is easy. The equations are dR/dt= BR- ARF and dF/dt= -DF+ ARF. The only non-linear terms are the products, ARF. Dropping those gives dR/dt= BR and dF/dt= -DF. Those are easy to solve.
To linearize about R= D/A, F= B/A, I would introduce the new variables, u= R- D/A, v= F- B/A. That way, the fixed point, R= D/A, F= B/A, becomes u= 0, v= 0. Of course since D/A and B/A are constants, du/dt= dR/dt and dv/dt= dF/dt. Also R= u+ D/A and F= v+ B/A.
The equations become du/dt= B(u+ D/A)+ A(u+ D/A)(v+ B/A)= Bu+ BD/A+ Auv+ Bu+ Dv+ BD/A and dv/dt= -D(v+ B/A)+ A(u+ D/A)(v+ B/A)= -Dv- BD/A+ Auv+ Bu+ Dv+ BD/A. The non-linear term is Auv and, since u and v are close to 0, can be dropped. That gives the two linear equations du/dt= Bu+ BD/A+ Bu+ Dv+ BD/A= 2Bu+ DV+ 2BD/A and dv/dt= -Dv- BD/A+ Bu+ Dv+ BD/A= Bu, not quite as easy to solve as the first two but still not very difficult.