Here is a result in the right direction, organizing the comments (some of which were unjustified, e.g., use of IBN). It applies to finite type von Neumann algebras and their regular rings, as well as all stably finite regular rings, and free rings too.

A (unital) ring is left *semihereditary* if every finitely generated left ideal is projective. This is a Morita invariant; in particular, this implies that every finitely generated submodule of a finitely generated free module is projective. Examples include von Neumann regular rings, rings all of whose matrix rings are Baer (thus including AW*-algebras), free rings, .... The class of semihereditary rings is closed under direct limits with one to one maps.

Recall that a ring is *stably finite* if all right invertible square matrices are invertible. This includes all AW*-algebras of finite type, their regular rings, free rings, and many others. The class of stably finite rings is closed under direct limits with one to one maps, and subrings.

**Proposition** Let $R$ be a left semihereditary stably finite ring. Then the condition $k_i \leq k_{i+1} + k_{i-1}$ is satisfied.

Proof. Let $P_1$ be the image of $M_i$ in $M_{i+1}$. As $M_{i+1}$ is free and $P_1$ is finitely generated (since $M_i$ is), we have that $P_1$ is projective. Hence $M_i \to P_1$ splits, and thus $M_i $ is isomorphic to $P_1 \oplus P_2$, where $P_2$ is the kernel of $M_i \to M_{i+1}$. We also have $P_1$ is the kernel of $M_{i+1} \to M_{i+2}$, so by the same argument, $P_1$ is a direct summand of $M_{i+1}$.

By exactness, $P_2$ is the image of $M_{i-1}$ in $M_i$, and freeness of $M_i$ (and $P_2$ finitely generated) implies $P_2$ is projective, and thus $P_2$ is isomorphic to a direct summand of $M_{i-1}$.

We have $M_i$ isomorphic to $P_1 \oplus P_2$, and $P_1$ is isomorphic to a direct summand of $M_{i+1}$ and $P_2$ is isomorphic to a direct summand of $M_{i-1}$. Hence $M_i$ is isomorphic to a direct summand of the free module $M_{i-1} \oplus M_{i+1}$. If $k_{i} > k_{i-1} + k_{i+1}$, we would obtain a contradiction to stable finiteness.\qed

Since semihereditariness is fairly strong, we should be able to deal with subrings, say $S \subset R$, where $R$ is at least semihereditary. Unfortunately, tensoring on the left with ${}_S R_S$ need not yield exactness of the sequence of free $R$-modules, $\cdots \to R \otimes_S M_{i} \to \cdots$ (although there may be ways around this). However, if $R_S$ is flat, then exactness is preserved. Hence:

**Corollary** Suppose $S\subset R$ is an inclusion of rings such that $R_S$ is flat, and $R$ is a left semihereditary stably finite ring. Then $k_i \leq k_{i+1} + k_{i-1}$.

It's been a long time since I did ring theory (25 years?).

**Edit** The corollary covers semiprime (right) Goldie rings, since their (classical) ring of quotients is flat. Also, if $K$ is a two-sided ideal such that $K^2 = K$, then $R/K$ is flat, covering some additional cases. In addition, a direct limit of rings with one to one maps each of which satisfies the inequality, also satisfies it.

$M_i$ is free on $k_i$generators (rather thanof rank $k_i$). In that case, stable finiteness (right inverses in matrix rings are left inverses) is likely sufficient, .... This is a slightly stronger hypothesis than invariant basis number. $\endgroup$ – David Handelman Dec 17 '16 at 0:33under the assumption that each $M_i$ is free on $k_i$ generators, in the proof that the inequality is correct for sf (or IBN) vn regular rings. $\endgroup$ – David Handelman Dec 17 '16 at 0:44