Let $F$ be a right exact additive covariant functor, and $L_iF( \cdot )$ its $i$-th left derived functor. I have read that if $F$ is also left exact, then $L_iF( M ) = 0$ for all $M$ and for all $i > 0$, because applying $F$ to a projective resolution preserves exactness.
The projective resolution of $M$ is something like: $$ \cdots \longrightarrow P_2 \longrightarrow P_1 \longrightarrow P_0 \longrightarrow M$$ and we ask ourselves whether $$ \cdots \longrightarrow FP_2 \longrightarrow FP_1 \longrightarrow FP_0 \longrightarrow 0$$ is exact at $FP_i$ for $i > 0$. As we look for exactness at $i > 0$ I don't understand where the hypothesis of $F$ being left exact comes into play. On the other hand I know this hypothesis is key, otherwise it would make little sense to study derived functors...
There is a slightly restricted (but entirely equivalent) formulation for derived functors that may be clarifying. A reference can be found in the appendix of Eisenbud's Commutative Algebra With a View Towards Algebraic Geometry (Proposition A3.17).
Given a right exact functor $F$, it takes $$0 \to A \to B \to C \to 0$$ and gives $$F(A) \to F(B) \to F(C).$$
Keeping in mind that the functor does not depend on our choice of resolution, we have that the first left-derived functor gives $L_1 F$ and a map $\delta:L_1F(C) \to F(A)$ so that $$L_1F(C) \to FA \to FB \to FC \to 0$$ is exact.
If the functor $F$ is exact, then $L_1F(C)=0$, by assumption.
More generally, the left derived funtors fit into the long exact sequence $$ \dots \to L_1F(A) \to L_1F(B) \to L_1F(C) \to L_0F(A) \to L_0F(B) \to L_0F(C) \to 0$$ and $L_0F(-)=F(-)$, which actually fully characterizes left derived functors.