When I try to compute $g^{ij}g^{pq}\nabla_{i,j}^2h_{pq}=\Delta tr_gh$, I need $g^{ij}=\delta^{ij},\nabla_i g_{kl}=0$,i.e, it's normal coordinate. But $g(t)$ will change with $t$, only for a $t_0$, I can use normal coordinate. So, whether $t=0$ is missed in the proof in picture below ?
The picture is from 59th page of this paper.
I think you are missing the essence of using normal coordinate, if you are able to derive $$g^{ij} g^{pq} \nabla^2_{ij} h_{pq} = \Delta \text{tr}_g h$$ using normal coordinate, then the above equation holds under ANY coordinate, as both sides of the equations are tensor fields. So to derive that equality, fix a $p$ and a $t$, then assume $g_t$ is the normal coordinate at $p$ and compute.