In this paper of Shulman, at page 4, he describes the (pseudo-)double category of rings and bimodules. This is supposed to be a double category with rings as objects, ring homomorphisms as tight 1-morphisms, bimodules as loose 1-morphisms and the two morphisms are "equivariant maps of bimodules". I do not understand the last bit (in scare quotes).
For instance, if we have $M$, a $R-S$ bimodule and $N$, a $R'-S'$ bimodule, what are the "equivariant" maps between $M$ and $N$? As far as I know, there are no relations/morphisms between $R,S,R',S'$.
Update: I am guessing that the data of an "equivariant map" $\psi: {_RM}_S \to {_{R'}M'}_{S'} $ is a triple $(\alpha:R \to R', \beta:S \to S', \psi:M \to M')$ where $\alpha, \beta$ are ring homomorphisms and $\psi$ is a map of bimodules with the induced action via the ring homomorphisms. Is this right?
Yes, your update is basically answering your question.
In a (pseudo) double category each cell has specific boundary arrows, two horizontal and two vertical.
It means that if $\psi$ is a cell, if its boundary is a bimodule $M:A-B$ on the top and a bimodule $M':A'-B'$ on the bottom, then its left side must be a ring homomorphism $\alpha:A\to A'$ and its right side $\beta:B\to B'$.
With these settings, the requirement for $\psi$ (beyond that it must be a homomorphism of the underlying Abelian groups) is simply: $$\psi(a\cdot m\cdot b)=\alpha(a)\cdot \psi(m)\cdot\beta(b)$$ for any $a\in A,\ b\in B,\ m\in M$.