Modules over commutative absolutely flat rings vs. vector spaces The theory of commutative absolutely flat rings (a.k.a. commutative von Neumann regular rings) is algebraic and furthermore it is the smallest variety containing all fields. Being a variety the category of (set-theoretic) models is arguably much better behaved than the category of fields.

How does linear algebra in modules over commutative absolutely flat rings compare
  to plain linear algebra, i.e. working with vector spaces? Is is it
  basically the same or are there some major differences?

Let's ask some more concrete questions:
Let $R$ be an commutative absolutely flat ring: Are finitely generated $R$-modules always free? Assuming the axiom of choice: Are all $R$-modules free?
 A: [OP]
"Let's ask some more concrete questions:
Let R
be an commutative absolutely flat ring: Are finitely generated R-modules always free? Assuming the axiom of choice: Are all R-modules free?"
[Answer]
Not necessarily.
Every nontrivial (unital) ring has a simple module, which is finitely generated, in fact 1-generated. R has a free simple left module iff R has no proper nontrivial left ideals iff R is a division ring. [The least obvious part of this claim is that if R has a free simple left module, then it has no proper nontrivial left ideals. You see this by observing that the only free module that could possibly be simple is the rank 1 free module, ${}_R R$. Then notice that this module is simple iff $R$ has no proper nontrivial left ideals.]
In the commutative case this shows that all (f.g.) $R$-modules are free iff $R$ is a field.
A: No. Simple example: let $k$ be field. The ring $R=k\times k$ is absolutely flat, and the first factor $k$, as an $R$-module, is the cyclic $R$-module generated by the idempotent $e=(1,0)$. It's not free, since $e$ is killed by the orthogonal idempotent $1-e=(0,1)$.
Actually one proves a commutative ring $R$ is absolutely flat if and only if its Krull dimension is $0$, and its localisations at maximal ideals are fields.
