Calculation of effective action in Einstein frame

I want to do the calculation for the effective action in Einstein frame (see Tong’s lecture notes Chapter 7.3), this action in string frame is given by
begin{align}label{exp1}
S=frac{1}{2 kappa_{0}^{2}} int d^{26} X sqrt{-G} e^{-2 Phi}left[mathcal{R}-frac{1}{12} H_{mu nu lambda} H^{mu nu lambda}+4 partial_{mu} Phi partial^{mu} Phiright].
end{align}
With a rescaling of the metric
$$tilde{G}_{mu nu}(X)=e^{-4 bar{Phi} /(D-2)} G_{mu nu}(X)$$
where $tilde{Phi} = Phi -Phi_0$, this should yield the following expression for the effective action:
$$ S_{Einstein}=frac{1}{2 kappa^{2}} int d^{26} X sqrt{-tilde{G}}left(tilde{mathcal{R}}-frac{1}{12} e^{-bar{Phi} / 3} H_{mu nu lambda} H^{mu nu lambda}-frac{1}{6} partial_{mu} tilde{Phi} partial^{mu} tilde{Phi}right), $$
for $D=26$.
Due to the change of the metric, the Ricci scalar also changes according to
$$
tilde{mathcal{R}}=e^{-2 omega}left(mathcal{R}-2(D-1) nabla^{2} omega-(D-2)(D-1) partial_{mu} omega partial^{mu} omegaright).
$$
with $omega=-frac{2 tilde{Phi}}{D-2}$. Now the introduction of $kappa^{2}=kappa_{0}^{2} e^{2 Phi_{0}}$ and taking into account that the second and third term in square brackets above depend on powers of the inverse metric, gives me the following intermediate result
begin{align}
S=frac{1}{2 kappa^{2}} int d^{D} X sqrt{-tilde{G}} e^{-2 omega}
left(
tilde{mathcal{R}}e^{2 omega} +
2(D-1) e^{2 omega} nabla^{2} omega +
(D-2)(D-1) e^{2 omega} partial_{mu} omega partial^{mu} omega +
qquad qquad + e^{6 omega} frac{1}{12} H_{mu nu lambda} H^{mu nu lambda} +
4 e^{2 omega} partial_{mu} tilde{Phi} partial^{mu} tilde{Phi}right),
end{align}
where I plugged
$$
mathcal{R} = e^{2 omega}tilde{mathcal{R}}+ 2(D-1) e^{2 omega} nabla^{2} omega + (D-2)(D-1) e^{2 omega}partial_{mu} omega partial^{mu} omega
$$
into the expression for the action in string frame.

My problem now is that I have no idea how to get the $-frac{1}{6} partial_{mu} tilde{Phi} partial^{mu} tilde{Phi} = -frac{4}{D-2} partial_{mu} tilde{Phi} partial^{mu} tilde{Phi}$ term as stated in the lecture notes. I think the Laplacian term in my intermediate result can be discarded, as it is a boundary term. Since the other terms do already match when $D=26$, I get
$$
-frac{4}{D-2} partial_{mu} tilde{Phi} partial^{mu} tilde{Phi} =
(D-2)(D-1) partial_{mu} omega partial^{mu} omega + 4 partial_{mu} tilde{Phi} partial^{mu} tilde{Phi}=(D-2)(D-1) partial_{mu} omega partial^{mu} omega + (D-2)^2 partial_{mu} omega partial^{mu} omega,
$$
which seems/is incorrect, unless I missed something trivial. Can someone point out to me, what I’m doing wrong here? Maybe I haven’t understood how rescaling the metric changes the action completely. Also I have noticed that if $(D-2)(D-1) partial_{mu} omega partial^{mu} omega$ would have opposite sign it would actually yield the correct result, but I don’t see why this should be the case. Help is much appreciated!