$\def\Ext{\operatorname{Ext}}$To prove your last statement, notice you can prove it by induction on the length of $N$, starting from the computation of $\Ext_S(k,S)$ which you can do trivially by using the Koszul resolution of $k$. (If you know about the Künneth formula for $\Ext$, this can be reduced to the computation of $\Ext_{k[X]}(k,k[X])$, with one variable, as $\Ext_{k[X_1,\dots,X_n]}(k,k[X_1,\dots,X_n])\cong\Ext_{k[X]}(k,k[X])^{\otimes n}$, but really, there is no need to do this, as the computation required is almost immediate)
If $N$ is of length $n>1$, and $0\subsetneq N'\subsetneq N$ is a non-zero proper submodule, we have a short exact sequence $0\to N'\to N\to N/N'\to 0$, and by induction we know that $\Ext^j(N',S)=\Ext^j(N/N',S)=0$ for all $j<r$, so the long exact sequence for Ext gives what the desired conclusion for $\Ext^j(N,S)$.
This is probably the official proof of the fact, I dunno.
To prove your first formula, notice that what we have just shown together with the fact that $\operatorname{gldim}S=r$ implies that $F=\Ext^r(\mathord-,S)$ is an exact functor on the category $\mathcal{A}$ of finite length graded modules, just as $G=\hom_k(\mathord-,k)=\hom_S(\mathord-,k)=\Ext^0_S(\mathord-,k)$ is.
There is a natural map $\Ext^0(M,k)\to \Ext^r(M,S)$ given by Yoneda product with the class of the $r$-extension given by the Koszul complex, which is an element of $\Ext^r(k,S)$. This map is an isomorphism when $M$ is simple, so general nonsense —the Five Lemma, in fact— implies that it is an isomorphism on all of $\mathcal A$.