A point in the $n$th space of $MSpin$ is an $n$-dimensional manifold equipped with a spin structure. In other words, it is a manifold equipped with a bundle of bimodules between the Clifford algebra of $\mathbb R^n$ and the Clifford algebra of $TM$. That bundle of bimodules is called the *spinor bundle* of $M$.

Similarly, a point in the $n$th space of $MSpin^c$ is an $n$-dimensional manifold equipped with a bundle of bimodules between the Clifford algebra of $\mathbb C^n$ and the Clifford algebra of $TM\otimes_{\mathbb R}\mathbb C$.

The map $MSpin\to MSpin^c$ is given by complexifying the bimodule.

A point in $n$th space of $KO$ (allow me to take non-connective $K$-theory - the result for connective $K$-theory then follows readily) is given by a real Hilbert space equipped with an action of $Cliff(\mathbb R^n)$, and an odd skew-adjoint clifford-linear Fredholm operator.

Similarly, a point in the $n$th space of $K$ is given by a complex Hilbert space with an action of $Cliff(\mathbb C^n)$, and an odd skew-adjoint clifford-linear Fredholm operator.

The map $KO\to K$ is again complexification.

The ABS orientation (which is not constructed in ABS)
sends a spin manifold $M$, now also equipped with a metric, to the Hilbert space of $L^2$ sections of the spinor bundle, equipped the the obvious $Cliff(\mathbb R^n)$-action. The Fredholm operator is the Dirac operator constructed from (the connection associated to) the metric and the $Cliff(TM)$-action.

The spin-c version is identical. It is then obvious by construction that the diagram you asked about is commutative.

*I should say that the above argument is completely hand-wavy...*

I actually don't know in which paper/textbook the ABS orientation is defined as a map of spectra (and I would like to know -- so if someone knows, please tell me). My guess is that, regardless of the approach taken, once you see the definition, it is completely obvious that the diagram is commutative.