For a proof of this result (and a more general version), there's a paper by Satoh [1] which has a lot of detail. The main idea is that for $TM$ to be Hermitian, the Nijenhuis tensor of $J^\prime$ needs to vanish. However, when you calculate the Nijenhuis tensor, you find that it vaishes if and only if the torsion and curvature of $D$ vanish (i.e. D is flat). See the equation on the bottom of page 8 of [1] for the exact formula.

For a Riemannian manifold, $g$ is flat iff the Levi-Civita connection is a flat connection, so I presume this is what the authors are using. For a more general connection $D$ (i.e. not the Levi-Civita connection), $(TM, J^\prime,g^D)$ is Kahler iff $(M,g,D)$ is a so-called Hessian manifold, which means that $D$ is flat and satisfies
$$(D_X g)(Y,Z)=(D_Y g)(X,Z)$$ for all vector fields $X,Y$ and $Z$. This relationship between the metric and connection is also known as Codazzi-coupling. The proof of this is also in that reference.

[1] *Satoh, Hiroyasu*, **Almost Hermitian structures on tangent bundles**, Suh, Young Jin (ed.) et al., Proceedings of the 11th international workshop on differential geometry, Taegu, Korea, November 9–11, 2006. Taegu: Kyungpook National University. 105-118 (2007). ZBL1125.53022.