To see why $K$-theory should be Brauer graded, it may help to see how, for a superalgebra $A$, the associated $K$-group only depends on the category $\mathcal{M}_A$ of finite dimensional $Z/2$-graded modules, which is endowed with a superstructure: it is enriched over the category of $Z/2$-graded vector spaces.

Given an object $X$ in such a $Z/2$-graded category $\mathcal{M}$, we can consider the set of odd degree endomorphisms $F: X^\pm\to X^\mp$. Say that $X$ is *trivial* if there exists such an $F$ for which $F^2=\mu\;\mathrm{Id}$, where $\mu$ is real and positive; I'll call $F$ a *trivialization* of $X$. Now define
$$K(\mathcal{M}) := (\text{Grothendieck group of $\mathcal{M}$} / \text{trivial objects}).$$
For modules over $\mathrm{Cliff}(n)$, this gives the expected group $K^n(\text{point})$ or $KO^n(\text{point})$, depending on your base field.
Note that $K(\mathcal{M})$ is clearly a Morita-equivalence invariant.

There's a pairing
$$K(\mathcal{M}_A) \otimes K(\mathcal{M}_B) \to K(\mathcal{M}_{A\otimes B}),$$
given by tensor product of modules. To see that this is well-defined, note that given odd endomorphisms $F_X\colon X^\pm\to X^\mp$ and $F_Y\colon Y^\pm\to Y^\mp$, you obtain an odd endomorphism
$$F=F_X\otimes \mathrm{Id}+\mathrm{Id}\otimes F_Y$$
of $X\otimes Y$, and that
$$F^2 = (F_X)^2\otimes \mathrm{Id} + \mathrm{Id} \otimes (F_Y)^2,$$
by the magic of sign conventions. Thus, if both $F_X$ and $F_Y$ are "trivializations" of $X$ and $Y$, or if just one of the two is a trivialization and the other is set to $0$, then $F$ is also a trivialization.

This suggests how $K$-theory should be Brauer-group graded, without explicitly invoking Clifford algebras.

This sort of thing seems "well-known", but I don't really know a reference. I learned this from Michael Joachim's paper "A Symmetric Ring Spectrum representing K-theory" (Topology, 2001), though he uses spaces of operators rather than finite dimensional modules.