Using Dunwoody's results on cohomological dimension to learn about a von Neumann regular group ring

I am not sure exactly what constitutes an answer to question 1, but here is a proof using the Dunwoody-Dicks stuff that if $G$ is a countable locally finite group, each of whose finite subgroups has order invertible in $R$, then the augmentation ideal of $RG$ is projective (recovering Kaplansky's result without using Connell).

Since $G$ is countable, we can write $G=\bigcup_{n\geq } G_n$ where $\{1\}=G_1\subset G_2\subset \cdots$ is a countable chain of finite subgroups (an infinite chain if $G$ is infinite, which is the interesting case) with orders invertible in $R$.

Let $T$ be the tree with vertex set $\coprod_{n\geq 1}G/G_n$ and we connect $gG_n$ to $gG_{n+1}$ by an edge for $g\in G$. Then $G$ acts on the tree $T$ in an obvious way, and the vertex stabilizers are conjugates of the $G_n$, and hence finite with orders invertible in $R$. Thus $G$ has a projective augmentation module by the result from Dunwoody-Dicks you cited.

The corresponding graph of groups is just a one-sided infinite ray with $G_i$ at vertex $i$ and edge $i$ and the edge groups include in the natural way ($G_i$ is mapped to $G_i$ by the identity and to $G_{i+1}$ by the inclusion) so the expression as a fundamental group of a graph of groups is the direct limit.

Having said that, this is most likely almost the same as Kaplansky's proof (I've never seen it) written in a geometric way. Going from an action on a tree with finite stabilizers of order invertible in $R$ to projective augmentation module is the easy direction. One just needs that the permutation module $R[G/H]$ is projective when $H$ is finite with order invertible in $R$, which is obvious since it is isomorphic to $RGe$ with $e$ the idempotent $e=\frac{1}{|H|}\sum_{h\in H}h$. Then one can use that the augmented simplicial chain complex of a tree is exact.

Going from projective augmentation module to the action on the tree is the hard direction and uses the Almost Stability Theorem, which is a fancy version of Stallings Ends Theorem.

I'm adding a new answer since my other answer was the converse.

Question 1:

I can `simplify' Connell's argument using Dunwoody-Dicks. Assume that $RG$ is von Neumann regular. Note that if $g\in G$, then $(1-g)r(1-g)=1-g$ for some $r\in R$. So $(1-g)(1- r(1-g))=0$. As $r(1-g)$ is in the augmentation ideal, it follows that $1-r(1-g)\neq 0$ and so $1-g$ is a left zero divisor. This implies $g$ has finite order by standard group ring arguments (in order for $a=ga$ we must have $g$ permutes the finite support of $a$ and hence has finite order). So $G$ is torsion.

If $G$ is countable, it has projective augmentation ideal (since countably generated left ideals in a von Neumann regular ring are projective) and so $G$ acts on a tree with finite stabilizers with order invertible $R$ by Dunwoody-Dicks and because the augmentation ideal is projective. Thus each finitely generated subgroup of $H$ also acts on a tree with finite stabilizers of order invertible in $R$ and we prove that $H$ is finite. But any finitely generated torsion group acing on a tree has a global fixed point (see Serre's book, where it is shown any finitely generated group of elliptic automorphisms has a global fixed point) and so $H$ is finite with order invertible in $R$. Alternatively, a finitely generated group acting on a tree with finite stabilizers has a finite index free subgroup by Bass-Serre theory, which must be trivial in our case since $H$ is a torsion group. So $H$ is finite and hence is a subgroup of a vertex stabilizer so its order is invertible in $R$. Thus $G$ is locally finite.