The Artin-Wedderburn theorem tells us that the maximal semisimple quotient is a product of matrix rings over finite division rings, one for each irreducible representation. Furthermore, every finite division ring is a field, and the unit group of any finite field is cyclic. The only nontrivial homomorphism from $S_3$ to a cyclic group is the sign homomorphism $S_3\to\mathbb{Z}/2$. It follows that any homomorphism from $\mathbb{Z}S_3$ to a finite field lands in the prime subfield (since elements of $S_3$ can only map to $\pm 1$).
So, writing $\mathbb{F}$ for either $\mathbb{F}_2$ or $\mathbb{F}_3$, the maximal semisimple quotient of $\mathbb{F}S_3$ is a product of matrix rings $M_n(K)$ for finite extensions $K$ of $\mathbb{F}$, one for each irreducible representation, and in all the cases where $n=1$ the $K$ is just $\mathbb{F}$. The only $1$-dimensional representations are the trivial representation and the sign representation, and the sign representation is the same as the trivial representation in the case $\mathbb{F}=\mathbb{F}_2$.
For $\mathbb{F}=\mathbb{F}_3$, dimension-counting now tells us there can be no more irreducible representations: the two $1$-dimensional representations take up $2$ dimensions of the semisimple quotient, and the Jacobson radical is nontrivial since it contains $\sum_{g\in S_3} g$, so there are at most $3$ dimensions left. Another irreducible representation would give a copy of $M_n(\mathbb{F}_{3^d})$ in the semisimple quotient for some $d$ and some $n>1$, which is impossible since there aren't enough dimensions left. We conclude that the two $1$-dimensional representations are the only irreducible representations for $\mathbb{F}=\mathbb{F}_3$, and so the maximal semisimple quotient is $\mathbb{F}_3\times\mathbb{F}_3$. The Jacobson radical is then the kernel of the map $\mathbb{F}_3S_3\to\mathbb{F}_3\times\mathbb{F}_3$; explicitly, it is the set of elements $\sum_{g\in S_3} a_g g$ such that $\sum a_g=0$ and $\sum a_g \sigma(g)=0$, where $\sigma(g)$ is the sign of $g$.
Over $\mathbb{F}_2$, on the other hand, there are up to $4$ dimensions left after accounting for the single $1$-dimensional representation and the fact that the Jacobson radical is nontrivial, so there might be a $2$-dimensional irreducible representation. To find one, note that there is a permutation representation of $S_3$ on $\mathbb{F}_2^3$, and this splits as a direct sum of a trivial subrepresentation (generated by $(1,1,1)$) and a $2$-dimensional subrepresentation (consisting of $(a,b,c)$ such that $a+b+c=0$). (Note that this splitting of the permutation representation doesn't happen over $\mathbb{F}_3$, since $(1,1,1)$ is contained in the latter $2$-dimensional subrepresentation.) This $2$-dimensional representation can easily be verified to be irreducible (for another way of seeing it, note that $\mathbb{F}_2^2\setminus\{0\}$ has three elements, and every permutation of them gives a linear map, so in fact $GL_2(\mathbb{F}_2)\cong S_3$).
So over $\mathbb{F}_2$, we conclude that there is the trivial representation and also this $2$-dimensional irreducible representation; counting dimensions, we now see that we have accounted for all $6$ dimensions of $\mathbb{F}_2S_3$. We conclude that the Jacobson radical is only $1$-dimensional (generated by $\sum_{g\in S_3} g$), and the quotient is $\mathbb{F}_2\times M_2(\mathbb{F}_2)$.
Best Answer
Yes, your approaches to 1-3 are fine.
For 4), you already know that you aren't going to have any luck with semisimple modules, so you need to use one that isn't semisimple. Why not start with $R$ itself?
$R$ itself is a $4$-dimensional $k$ algebra, so we'll use that as $M$.
Then just let $N$ be a semisimple module of length $4$ (four copies of the unique simple module.)
Obviously $M\ncong N$, but they have the same $k$-dimension.
Answering 4) above does that, right?