= Solution
A <linear connection on a manifold> is a rule $(X,Y)\mapsto\nabla_XY$ for differentiating <vector fields> which is additive, is $C^\infty$-linear in $X$, is real-linear in $Y$, and obeys $\nabla_X(fY)=X(f)Y+f\nabla_XY$. It extends to <covector fields> and arbitrary <tensor fields> by the <Leibniz rule> and compatibility with contraction. In coordinates,
$$
\nabla_aU^d=\partial_aU^d+\Gamma^d{}_{ac}U^c,\qquad
\nabla_a\omega_c=\partial_a\omega_c-\Gamma^d{}_{ac}\omega_d,\qquad
\nabla_af=\partial_af.
$$
This is the <covariant derivative>; the <affine connection> terms correct the changes of the coordinate <basis>. A symmetric <affine connection> has zero <torsion tensor>, $T(X,Y)=\nabla_XY-\nabla_YX-[X,Y]=0$, or $\Gamma^c{}_{ab}=\Gamma^c{}_{ba}$ in a coordinate <basis>. Symmetry does not imply <metric compatibility>; no metric is required for this question.
Additivity of each <covariant derivative> immediately gives additivity of $\Delta_{ab}$. Expanding the second derivative of a <tensor product> gives
$$
\begin{aligned}
\nabla_a\nabla_b(Q\otimes S)={}&(\nabla_a\nabla_bQ)\otimes S+(\nabla_bQ)\otimes\nabla_aS\\
&+(\nabla_aQ)\otimes\nabla_bS+Q\otimes\nabla_a\nabla_bS.
\end{aligned}
$$
Subtract the expression with $a,b$ exchanged. The two mixed terms cancel, proving that the <curvature commutator is a tensor derivation>:
$$
\boxed{\Delta_{ab}(Q\otimes S)=(\Delta_{ab}Q)\otimes S+Q\otimes\Delta_{ab}S.}
$$
For a scalar, $\nabla_a\nabla_bf=\partial_a\partial_bf-\Gamma^c{}_{ab}\partial_cf$. Commuting partial derivatives and using the symmetric <affine connection> gives $\Delta_{ab}f=0$. Consequently $\Delta_{ab}(fU)=f\Delta_{ab}U$: the <commutator> is linear over smooth functions, not merely over constant scalars. Write $U=U^ce_c$ in a local <basis>. The <tensor>-product rule gives $\Delta_{ab}U=U^c\Delta_{ab}e_c$, showing that its value at a point depends only on $U$ at that point. Thus it defines the <Riemann curvature tensor> by
$$
\boxed{\Delta_{ab}U^d=R_{abc}{}^dU^c,\qquad
R_{abc}{}^d=\partial_a\Gamma^d{}_{bc}-\partial_b\Gamma^d{}_{ac}+\Gamma^d{}_{ae}\Gamma^e{}_{bc}-\Gamma^d{}_{be}\Gamma^e{}_{ac}.}
$$
The antisymmetrized second <covariant derivative> is itself tensorial in the derivative indices, so this pointwise map transforms as a <tensor>. The formula or the defining <commutator> gives $R_{abc}{}^d=-R_{bac}{}^d$, hence $R_{abc}{}^d=R_{[ab]c}{}^d$ with normalized antisymmetrization.
Apply the <commutator> to the scalar contraction $\omega_cU^c$. Its action vanishes, so
$$
0=(\Delta_{ab}\omega_c)U^c+\omega_cR_{abd}{}^cU^d.
$$
Since $U$ is arbitrary, $\Delta_{ab}\omega_c=-R_{abc}{}^d\omega_d$. Repeated use of the <tensor>-product rule gives one plus <curvature> action for an upper index and one minus action for each lower index. In particular,
$$
\boxed{\Delta_{ab}S^c{}_{de}=R_{abf}{}^cS^f{}_{de}-R_{abd}{}^fS^c{}_{fe}-R_{abe}{}^fS^c{}_{df}.}
$$
The type notation in the PDF means one contravariant and two covariant indices, not a fractional <tensor> order.
For the <first Bianchi identity>, put $\omega_c=\nabla_cf$ and $H_{bc}=\nabla_b\nabla_cf$. The <symmetric Hessian of a scalar field> satisfies $H_{bc}=H_{cb}$. Therefore the six third-derivative terms cancel in
$$
\Delta_{ab}\omega_c+\Delta_{bc}\omega_a+\Delta_{ca}\omega_b
=\nabla_aH_{bc}-\nabla_bH_{ac}+\nabla_bH_{ca}-\nabla_cH_{ba}+\nabla_cH_{ab}-\nabla_aH_{cb}=0.
$$
Substituting the covector <curvature> action gives $(R_{abc}{}^d+R_{bca}{}^d+R_{cab}{}^d)\nabla_df=0$. At a chosen point the <gradient> of a smooth scalar can be any covector, so
$$
\boxed{R_{abc}{}^d+R_{bca}{}^d+R_{cab}{}^d=0,\qquad R_{[abc]}{}^d=0.}
$$
The second form follows from the already proved first-pair antisymmetry.
For the <second Bianchi identity>, the same third-derivative cancellation can be organized as an operator identity. On vector-valued components set $D_a=\partial_a+\Gamma_a$, with $(\Gamma_a)^e{}_d=\Gamma^e{}_{ad}$. Although a full second <covariant derivative> includes a term $-\Gamma^f{}_{ab}D_f$ for its lower derivative index, that term cancels on antisymmetrization because the <affine connection> is symmetric. Thus $[D_b,D_c]=\mathcal R_{bc}$, where $(\mathcal R_{bc})^e{}_d=R_{bcd}{}^e$ acts by multiplication. Expanding the three nested operator <commutators> makes all six triple products cancel:
$$
[D_a,[D_b,D_c]]+[D_b,[D_c,D_a]]+[D_c,[D_a,D_b]]=0.
$$
Equivalently, the cyclic sum of $\partial_a\mathcal R_{bc}+[\Gamma_a,\mathcal R_{bc}]$ vanishes. These are the <covariant derivative> terms on the upper and last lower <curvature> indices. The two remaining lower indices add $-\Gamma^f{}_{ab}R_{fcd}{}^e-\Gamma^f{}_{ac}R_{bfd}{}^e$. In the cyclic sum these additions cancel in pairs, using $\Gamma^f{}_{ab}=\Gamma^f{}_{ba}$ and $R_{fcd}{}^e=-R_{cfd}{}^e$. Hence the <Bianchi identities from covariant derivative commutators> give
$$
\boxed{\nabla_aR_{bcd}{}^e+\nabla_bR_{cad}{}^e+\nabla_cR_{abd}{}^e=0,\qquad\nabla_{[a}R_{bc]d}{}^e=0.}
$$
Both identities here apply to any torsion-free linear <affine connection>; the second derivation does not require choosing a metric <affine connection>.
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