The commutator of covariant derivatives on a vector field and its contracted component identity

(The uncontracted identity.) Both sides of (a) are components of smooth tensor fields, so it suffices to verify the identity at an arbitrary point $p$. Take a point-normal chart at $p$ as in [L3]. There $A^a{}_c(p)={\mathrm{\partial }}_c{Z}^a(p)$. From [L2] and $\mathrm{\Gamma }(p)=0$, $B^a{}_{bc}(p)={\mathrm{\partial }}_b{\mathrm{\partial }}_c{Z}^a(p)+{\mathrm{\partial }}_b{\mathrm{\Gamma }}^a{}_{cd}(p)Z^d(p)$. The second derivatives of $Z$ cancel after antisymmetrizing, giving $B^a{}_{bc}(p)-B^a{}_{cb}(p)=({\mathrm{\partial }}_b{\mathrm{\Gamma }}^a{}_{cd}(p)-{\mathrm{\partial }}_c{\mathrm{\Gamma }}^a{}_{bd}(p))Z^d(p)$. By [L1], the coordinate formula for $R({\mathrm{\partial }}_b,{\mathrm{\partial }}_c){\mathrm{\partial }}_d$ is $R^a{}_{dbc}={\mathrm{\partial }}_b{\mathrm{\Gamma }}^a{}_{cd}-{\mathrm{\partial }}_c{\mathrm{\Gamma }}^a{}_{bd}+{\mathrm{\Gamma }}^e{}_{cd}{\mathrm{\Gamma }}^a{}_{be}-{\mathrm{\Gamma }}^e{}_{bd}{\mathrm{\Gamma }}^a{}_{ce}$. At $p$ its quadratic terms vanish, so its coefficient is exactly the one above. Thus $B^a{}_{bc}-B^a{}_{cb}=R^a{}_{dbc}{Z}^d$ at $p$ and hence everywhere, proving (a).

(The contracted identity.) Set the first index equal to the second lower index in (a) and sum, as the Einstein convention of Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model prescribes: $B^a{}_{ab}-B^a{}_{ba}=R^a{}_{dab}{Z}^d$. By [L2], ${\mathrm{Ric}}_{db}=R^a{}_{dab}$ (relabel ${\mathrm{Ric}}_{bd}=R^a{}_{bad}$ by exchanging the free indices), and ${\mathrm{Ric}}_{db}={\mathrm{Ric}}_{bd}$, so $B^a{}_{ab}-B^a{}_{ba}={\mathrm{Ric}}_{bd}{Z}^d$, which is (b).

(The operational form.) Let ${\theta }_Z:={\mathrm{\nabla }}_a{Z}^a=A^a{}_a$ be the divergence of $Z$, a smooth function. By [L2] the covariant derivative of a contraction is the contraction of the derivative, so ${\mathrm{\nabla }}_c{\theta }_Z={\mathrm{\nabla }}_c(A^a{}_a)=B^a{}_{ca}$. By (b) with the free index $b$ renamed to $c$, $B^a{}_{ac}-B^a{}_{ca}={\mathrm{Ric}}_{cd}{Z}^d$, hence $B^a{}_{ca}=B^a{}_{ac}-{\mathrm{Ric}}_{cd}{Z}^d={\mathrm{\nabla }}_a{A}^a{}_c-{\mathrm{Ric}}_{cd}{Z}^d={\mathrm{\nabla }}_a({\mathrm{\nabla }}_c{Z}^a)-{\mathrm{Ric}}_{cd}{Z}^d$, where the last step uses the definition $B^a{}_{ac}={\mathrm{\nabla }}_a{A}^a{}_c$ and $A^a{}_c={\mathrm{\nabla }}_c{Z}^a$. Multiplying by $V^c$ and using the Leibniz rule for the contracted product, $V^c{\mathrm{\nabla }}_a({\mathrm{\nabla }}_c{Z}^a)={\mathrm{\nabla }}_a\left(V^c{\mathrm{\nabla }}_c{Z}^a\right)-({\mathrm{\nabla }}_a{V}^c)({\mathrm{\nabla }}_c{Z}^a)$. Substituting this and the Ricci term into $V^c{\mathrm{\nabla }}_c{\theta }_Z$ gives the displayed identity of (c), which completes the proof.