 Solution

Use the convention consistent with the supplied fluid transformations: $\delta g_{ij}=a^2{h}_{ij}$, so the synchronous spatial metric is $g_{ij}=-a^2{\delta }_{ij}+a^2{h}_{ij}$. For a passive coordinate displacement ${\xi }^0=T{e}^{i\mathbf{k}\cdot \mathbf{x}}$, ${\xi }^i=i{\hat{k}}^{i}L{e}^{i\mathbf{k}\cdot \mathbf{x}}$, the metric perturbation changes by ${\widetilde{\delta g}}_{\mu \nu }=\delta g_{\mu \nu }-({\mathcal{L}}_{\xi }\bar{g}{)}_{\mu \nu }$. Computing its components givesThe Newtonian gauge has no mixed component and no traceless spatial part. ThusMatching ${\widetilde{\delta g}}_{00}=2a^2\Phi$ and ${\widetilde{\delta g}}_{ij}=2a^2\Psi {\delta }_{ij}$ givesThis is the synchronous-to-Newtonian transformation with positive spatial perturbations. The original PDF uses $i\hat{\mathbf{k}}L$ in the spatial displacement; the converted TeX loses the hat. Using $i\mathbf{k}L$ would require $L=h_S/(2k^2)$ instead. The nonzero modes $k>0$ are understood.