For Question 2, The central extension $\tilde{S}_4$ is certainly a subgroup of $\mathrm{GL}_2(\mathbf{Z}[\sqrt{-2}]) \subset \mathrm{GL}_2(\mathbf{C})$. The image of the determinant is $\pm 1$. The image of $\tilde{S}_4$ in
$$\mathrm{GL}_2(\mathbf{Z}[\sqrt{-2}]/2) = \mathrm{GL}_2(\mathbf{F}_2[x]/x^2)$$
is $S_4$, and all the elements have determinant one. It's easy to see that the central element
$$\left( \begin{matrix} -1 & 0 \\\ 0 & -1 \end{matrix} \right)$$
lies in the kernel, so it suffices to note that nothing else does. Yet it's obvious that the map surjects onto $\mathrm{GL}_2(\mathbf{F}_2) = S_3$, and (from the character table) the image is larger than $S_3$, so the image is $S_4$.
For Question 1, if $G$ injects into $\mathrm{SL}_2(R)$ for some $R$ then it injects into such a ring where $R$ is Artinian. Here is the proof.
EDIT: Step 0. (This was in my head, but I forgot to mention it, as Kevin reminds me in the comments). One may replace $R$ by the subring generated by the images of the entries of $g-1$ for all $g \in G$, and hence assume that $R$ is finitely generated over $\mathbf{Z}$ and hence Noetherian. (The Krull intersection thm requires a Noetherian hypothesis.)
Step 1. If $x$ is a non-zero element of $R$, then there exists a maximal ideal $\mathfrak{m}$ of $R$ such that $x$ is non-zero in $R/\mathfrak{m}^k$ for some $k$. Proof: Let $\mathfrak{m}$ be some maximal ideal containing the annihilator of $x$. Then $x$ is non-zero in the localization $R_{\mathfrak{m}}$, and thus $x$ is non-zero in $R/{\mathfrak{m}^k}$ by the Krull intersection theorem.
Step 2. If $x_1, \ldots, x_n$ are non-zero elements of $R$, there exists an ideal $I$ such that each $x_i$ is non-zero in $R/I$ and $R/I$ is Artinian. Proof: Apply Step 1 to each $x_i$, and let $I = \bigcap \mathfrak{m}^{k_i}_i$.
Step 3. Suppose that $G$ has $n$ non-trivial elements. Let $x_1, \ldots, x_n$ denote
a non-zero entry in the matrix $g - 1$ for each element of $g$. Apply Step 2 to deduce that
$g$ is not the identity in $R/I$ for some Artinian quotient for all non-zero $g \in G$.
Remark: If $G$ is simple, then $G$ is actually a subgroup of $\mathrm{SL}(k)$ for some field $k$. Proof: Artinian rings are semi-local, so $G$ is a
subgroup of $\bigoplus_{i=1}^{n} \mathrm{SL}(A_i)$ for Artinian rings $A_i$. Since $G$ is simple, it must be a subgroup of $\mathrm{SL}(A)$ for one such $A$. This latter group is filtered by the groups $\mathrm{SL}(k)$ and copies of $M_0(k)$ (trace zero matrices). Since the latter is abelian and $G$ is simple, we are done.
It's easy to find examples of groups which are not subgroups of $\mathrm{SL}_n(k)$ for all fields $k$ and some fixed integer $n$.
Best Answer
In every finite $p$-group $G$ of order $p^n$ there is a normal subgroup $N_k$ of order $p^k$ for any $k\le n$. The proof is by induction on $k$. For $k=1$ take a cyclic subgroup of order $p$ in the center, for the step, take a cyclic central subgroup in $G/N_k$ and its preimage in $G$. So if you do not assume that $A\cap B=\{1\}$, one can't say anything specific about the orders. If you assume that $A\cap B=\{1\}$, i.e. your product is semi-direct, then some information (not much, though) can be deduced.
If $|A|=p$ and $A$ is normal, then $A$ is central because in a nilpotent group every non-trivial normal subgroup intersects the center non-trivially. In that case either $B=G$ or $A\cap B=\{1\}$ and $G=A\times B$.
If $|A|=p^k, k\ge 2$, then again you can take a central cyclic of order $p$ subgroup $T$ of $A$, and you will have $G/T=(A/T)(BT/T)$. This reduces your problem to the same problem for a smaller group $G/T$.