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.

To prove that a right-adjoint additive functor is left-exact, use the adjunction isomorphism to build a commutative rectangle from some exact sequence and then use the fact that the Yoneda embedding reflects exactness.

I'm unsure about the more general setting. Although in that case you would need to prove that $\mathcal{F}$ preserves finite limits, which includes preserving terminal objects, i.e. $\mathcal{F}(0)=0$.