= Solution
Use the convention consistent with the supplied fluid transformations: $\delta g_{ij}=a^2h_{ij}$, so the synchronous spatial metric is $g_{ij}=-a^2\delta_{ij}+a^2h_{ij}$. For a passive coordinate displacement $\xi^0=T e^{i\mathbf k\cdot\mathbf x}$, $\xi^i=i\hat k^iL e^{i\mathbf k\cdot\mathbf x}$, the <metric perturbation> changes by $\widetilde{\delta g}_{\mu\nu}=\delta g_{\mu\nu}-(\mathcal L_\xi\bar g)_{\mu\nu}$. Computing its components gives
$$
\widetilde{\delta g}_{00}=-2a^2(\mathcal HT+T'),\quad\widetilde{\delta g}_{0i}=ia^2k_i(L'/k-T),
$$
$$
\frac{\widetilde{\delta g}_{ij}}{a^2}=\left[\frac{h-h_S}{3}+2\mathcal HT\right]\delta_{ij}+(h_S-2kL)\hat k_i\hat k_j.
$$
The <Newtonian gauge> has no mixed component and no traceless spatial part. Thus
$$
\boxed{L=\frac{h_S}{2k},\qquad T=\frac{L'}k=\frac{h_S'}{2k^2}}.
$$
Matching $\widetilde{\delta g}_{00}=2a^2\Phi$ and $\widetilde{\delta g}_{ij}=2a^2\Psi\delta_{ij}$ gives
$$
\boxed{\Phi=-\mathcal HT-T'=-\frac{\mathcal Hh_S'+h_S''}{2k^2},\qquad\Psi=\frac{h-h_S}{6}+\frac{\mathcal Hh_S'}{2k^2}}.
$$
This 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 $i\mathbf kL$ would require $L=h_S/(2k^2)$ instead. The nonzero modes $k>0$ are understood.
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