I am not sure what your goal is with this question, but I think there is an inherent problem with your set up. $RF$ stands for a functor defined from $F$ and not just a functor on the derived category. If you consider the derived category only as a triangulated category via a forgetful functor then you are also forgetting $F$, so there is no reason to still denote your functor by $RF$. On the other hand if you remember the derived category structure or at least you still deal with complexes and so you can take cohomology of these complexes, then $RF$ induces $R^iF$ and you have the condition that $R^iF=0$ for $i<0$ and $R^0F=F$. In other words, the functor $RF$ is such that the leftmost cohomology of it is $F$ and all the interesting stuff is on the right. Homework: do the same for $LG$.

It is important that $RF$ is the right derived functor of $F$. It can happen that $RF=LG$, but then $F$ and $G$ are different. Typically $G=R^nF$ such that $R^mF=0$ for $m>n$.

I am not sure what your goal is with this question, but I think there is an inherent problem with your set up. $RF$ stands for a functor derived from $F$ and not just a functor on the derived category. If you consider the derived category only as a triangulated category via a forgetful functor then you are also forgetting $F$, so there is no reason to still denote your functor by $RF$. On the other hand if you remember the derived category structure or at least you still deal with complexes and so you can take cohomology of these complexes, then $RF$ induces $R^iF$ and you have the condition that $R^iF=0$ for $i<0$ and $R^0F=F$. In other words, the functor $RF$ is such that the leftmost non-zero cohomology of it is $F$ and all the interesting stuff is on the right. Homework: do the same for $LG$.

It is important that $RF$ is the right derived functor of $F$. It can happen that $RF=LG$, but then $F$ and $G$ are different. Typically $G=R^nF$ such that $R^mF=0$ for $m>n$.