= Solution
Use the round unit sphere with its complex orientation and unique <spin structure>. Its chiral <spinor bundles> are $S^+=\mathcal O(-1)$ and $S^-=\Lambda^{0,1}\otimes\mathcal O(-1)$, the latter a smooth line bundle of degree $+1$. The <Dirac operator> is
$$
D=c\circ\nabla^S=\begin{pmatrix}0&D^-\\D^+&0\end{pmatrix},\qquad
D^+=\sqrt2\,\overline\partial_{\mathcal O(-1)},\quad D^-=(D^+)^*.
$$
Here <Clifford multiplication> has $c(v)^2=-|v|^2$, and $D$ acts self-adjointly on the L2 spinors with first-order Sobolev domain. The geometric spin connection is invariant under the lifted $SU(2)$ action, hence $D$ is equivariant.
We justify the spectral properties needed for its phase. The <Peter-Weyl theorem> and <Frobenius reciprocity for compact groups> give the decomposition of sections of an isotropy-weight-$q$ line bundle: $E_n$ occurs once exactly when $n\geq|q|$ and $n\equiv q\pmod2$, because its torus weights are $n,n-2,\ldots,-n$. The spinor isotropy weights are $+1,-1$, so each chirality contains $E_1,E_3,E_5,\ldots$, each once. The horizontal connection Laplacian is the total Casimir minus the vertical generator square, $\nabla^*\nabla=C-1/4$ on these weight-one bundles. Expanding the Dirac square in a normal orthonormal frame gives the diagonal connection-Laplacian terms and the off-diagonal curvature contraction. The Clifford relations contract spin curvature to $\operatorname{Scal}/4$, the <Lichnerowicz formula for a twisted Dirac operator> with no twisting bundle. Since the scalar curvature here is two,
$$
D^2=\nabla^*\nabla+\tfrac12=C+\tfrac14,
\qquad D^2|_{E_n}=\frac{(n+1)^2}{4}I.
$$
Thus there is no kernel, the <eigenvalues> of $D$ are $\pm1,\pm2,\ldots$ with finite multiplicities, and the resolvent is compact by spectral truncation. These round-sphere spectral facts agree with the explicit calculation in https://arxiv.org/abs/hep-th/0212134[Abrikosov's paper].
Define the <chiral phase of the round-sphere Dirac operator>
$$
\boxed{V=D^+(D^-D^+)^{-1/2}:H_+\longrightarrow H_-,\qquad H_\pm=L^2(S^\pm).}
$$
It is unitary because both chiral kernels vanish; equivariance follows from that of $D$. The full phase is $F=D|D|^{-1}=\begin{pmatrix}0&V^*\\V&0\end{pmatrix}$.
For a smooth scalar $f$, the Leibniz rule gives $[D,M_f]=c(df)$, a bounded bundle endomorphism. The spectral integral for the sign operator gives, in the strong topology,
$$
F=\frac2\pi\int_0^\infty D(D^2+t^2)^{-1}\,dt.
$$
Write $R_\pm=(D\mp it)^{-1}$. Since $2D(D^2+t^2)^{-1}=R_++R_-$ and $[R_\pm,M_f]=-R_\pm[D,M_f]R_\pm$, the commutator integral is norm convergent: its integrands are compact and bounded by $\|[D,M_f]\|/(1+t^2)$. A norm limit of their integrals is compact. Thus $[F,M_f]$ is compact, and its off-diagonal blocks say $VM_f^+-M_f^-V$ is compact. Uniform approximation of continuous $f$ by smooth functions extends this to all $f\in C(S^2)$, using the bound $2\|f-g\|_\infty$ for the change of commutator.
\b[The chiral phase is an equivariant unitary which intertwines the two multiplication representations modulo compacts.] This specifies the grading implicit in the subsequent index question; treating the full self-adjoint phase as an endomorphism would yield a different, zero, compressed index.
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