= Solution
The paper uses $T_{\mu\nu}{}^\lambda=\Gamma_{\nu\mu}{}^\lambda-\Gamma_{\mu\nu}{}^\lambda$, the negative of a common <torsion tensor> convention. Keep that sign throughout. Also, $\nabla_\mu\nabla_\nu\varphi$ denotes the covariant <Hessian>, so the second derivative acts on the lower index $\nu$ of the <covector> $\nabla_\nu\varphi$; it is not just a composition of two directional derivatives on a scalar.
Since the <covariant derivative> of a <scalar field> is its ordinary derivative,
$$
\nabla_\mu\nabla_\nu\varphi
=\partial_\mu\partial_\nu\varphi-\Gamma_{\mu\nu}{}^\lambda\partial_\lambda\varphi.
$$
Commuting the ordinary partial derivatives therefore gives
$$
\boxed{(\nabla_\mu\nabla_\nu-\nabla_\nu\nabla_\mu)\varphi
=T_{\mu\nu}{}^\lambda\nabla_\lambda\varphi.}
$$
For a <vector field>, write $D_\nu X^\lambda=\partial_\nu X^\lambda+\Gamma_{\nu\rho}{}^\lambda X^\rho$. This is a <tensor> with one covariant and one contravariant index. Hence
$$
\nabla_\mu\nabla_\nu X^\lambda
=\partial_\mu(D_\nu X^\lambda)+\Gamma_{\mu\sigma}{}^\lambda D_\nu X^\sigma
-\Gamma_{\mu\nu}{}^\sigma D_\sigma X^\lambda.
$$
Subtracting the expression with $\mu,\nu$ exchanged, the first two terms combine to
$$
\left(\partial_\mu\Gamma_{\nu\rho}{}^\lambda
-\partial_\nu\Gamma_{\mu\rho}{}^\lambda
+\Gamma_{\mu\sigma}{}^\lambda\Gamma_{\nu\rho}{}^\sigma
-\Gamma_{\nu\sigma}{}^\lambda\Gamma_{\mu\rho}{}^\sigma\right)X^\rho.
$$
The coefficients of first derivatives of $X$ in those two terms cancel. The last terms give $(\Gamma_{\nu\mu}{}^\sigma-\Gamma_{\mu\nu}{}^\sigma)D_\sigma X^\lambda$. Identifying the <curvature> coefficient yields
$$
\boxed{[\nabla_\mu,\nabla_\nu]X^\lambda
=R_{\mu\nu\rho}{}^\lambda X^\rho
+T_{\mu\nu}{}^\rho\nabla_\rho X^\lambda.}
$$
Here the bracket denotes antisymmetrization of the <tensorial> second <covariant derivative>, consistently with the preceding calculation.
To check tensoriality directly, antisymmetrize the supplied transformation law of the <connection coefficients> in the two lower indices. The inhomogeneous term vanishes because mixed coordinate second derivatives are symmetric. The remaining transformation law is
$$
T_{\nu\lambda}{}^\mu
=\frac{\partial x^\mu}{\partial x^{\mu'}}
\frac{\partial x^{\nu'}}{\partial x^\nu}
\frac{\partial x^{\lambda'}}{\partial x^\lambda}
T_{\nu'\lambda'}{}^{\mu'},
$$
which is precisely that of a $(1,2)$ <tensor>. Intrinsically, with the paper's sign,
$$
\boxed{T(X,Y)=\nabla_YX-\nabla_XY+[X,Y].}
$$
The derivative terms involving the components of $X,Y$ cancel against their <Lie bracket>, leaving $(\Gamma_{\nu\mu}{}^\lambda-\Gamma_{\mu\nu}{}^\lambda)X^\mu Y^\nu$. This also shows that the expression is linear over smooth functions in each argument.
Finally take the <torsion-free connection> and the commuting parameter fields $Y,Z$ along the <geodesic> family. Vanishing torsion gives $\nabla_YZ=\nabla_ZY$, while affine <geodesic> parametrization gives $\nabla_YY=0$. The <curvature> operator in the specified convention is
$$
R(Y,Z)W=\nabla_Y\nabla_ZW-\nabla_Z\nabla_YW-\nabla_{[Y,Z]}W.
$$
Therefore
$$
\nabla_Y\nabla_YZ
=\nabla_Y\nabla_ZY
=R(Y,Z)Y+\nabla_Z\nabla_YY+\nabla_{[Y,Z]}Y
=R(Y,Z)Y.
$$
In components this is the <geodesic deviation> equation
$$
\boxed{\frac{D^2Z^\mu}{d\tau^2}
=R_{\nu\lambda\rho}{}^\mu Y^\nu Z^\lambda Y^\rho.}
$$
The connecting field represents infinitesimal separation within the family, so this is a linear equation for relative free-fall acceleration, rather than an exact coordinate difference for widely separated worldlines.
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