= Solution
For a <torsion-free connection>, the covariant Hessian of a scalar is symmetric, so $[\nabla_a,\nabla_b]f=0$. Moreover, a commutator of two covariant derivatives acts as a derivation on tensor products. Applying the <Leibniz rule> twice, the terms containing one derivative on each factor cancel between the two orders:
$$
[\nabla_a,\nabla_b](u\otimes v)=([\nabla_a,\nabla_b]u)\otimes v+u\otimes([\nabla_a,\nabla_b]v).
$$
Projecting with the parallel <symplectic form> preserves this identity for the <chiral spinor curvature operator>. Since its scalar projection obeys $\Delta_{AB}f=0$, we obtain
$$
\boxed{\Delta_{AB}(f\alpha^C\beta^{C'})=f\left(\beta^{C'}\Delta_{AB}\alpha^C+\alpha^C\Delta_{AB}\beta^{C'}\right).}
$$
Thus the <chiral spinor curvature operator> is algebraic over scalar functions even though its displayed expression uses second derivatives. If <torsion tensor> is allowed, $[\nabla_a,\nabla_b]f=-T^c{}_{ab}\partial_cf$ need not vanish, explaining why metric compatibility alone would not prove the claim.
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