Right-adjoint functors are left-exact?

As Martin Brandenburg mentioned, this holds in a much more general context:

Let $C,D$ be categories and $F\colon C\rightarrow D$, $G\colon D\rightarrow C$ functors, such that $F$ is left adjoint to $G$. Then $F$ preserves all colimits and $G$ preserves all limits. Especially G preserves kernels and therefore is left-exact, whenever you can talk about exactness. (Dually: F preservers cokernels, so it's right-exact.)

This isn't hard to prove: You can prove by hand (checking the universal property), that covariant homfunctors $Hom(A,\_)$ preserve finite limits and then you can do the following by using the adjointness and the preservation of limits of hom-functors:

$$Hom(A,G(lim(X_i)))\cong Hom(F(A),lim(X_i))\cong lim(Hom(F(A),X_i))\cong lim(Hom(A,G(X_i)))\cong Hom(A,lim(G(X_i)))$$

All these isomorphisms are natural in A, so we get $Hom(\_,G(lim(X_i))\cong Hom(\_,lim(G(X_i)))$ and with the fact, that the Yoneda embedding is an embedding $G(lim(X_i))\cong lim(G(X_i))$. Dually this works for rightadjoints and colimits.

As other have stated, there are lots of good properties for adjoint functors. I will just prove your $F(0)=0$ statement. Let $G$ be left adjoint par to $F$. Then for all $S$-modules $X$ and $R$-modules $Y$ we have bijection $Hom_R(GX,Y) \simeq Hom_S(X,FY)$. Put $Y=0$ and $X=F0$. So we get:

$$Hom_S(F0,F0) \simeq Hom_R(GF0,0) \simeq \{0\}.$$

Now, if $F0\neq 0$, then there would be at least two different S-module-homomorphisms $F0 \to F0$ (namely, identity and zero-homomorphism), which would be a contradiction.