Let $\Gamma$ be a $(n-1)$-dimensional compact hypersurface (with whatever smoothness is required). Is it true that $$C^\infty([0,T]\times \Gamma) \subset C^\infty([0,T];H^1(\Gamma))$$ holds? I'm not sure that it is since if $\varphi \in C^\infty([0,T]\times \Gamma)$ and $t_n \to t$ we would need for example $$\int_{\Gamma}|\varphi(t_n)-\varphi(t)|^2 \to 0$$ and I don't see that this holds. But the domain is compact and we do have uniform convergence...
Is the inclusion dense if it is an inclusion?
It is true and it is dense.
If $f\in C^\infty([0,T]\times \Gamma)$ then $\partial^n_t u$ clearly is $C^\infty$ is $t$ and belongs to $H^1(\Gamma)$, for all $t\in[0,T]$, as it is sufficiently smooth with respect to the variables which paprametrize $\Gamma$.
Also, as $\Gamma$ is compact and smooth, $H^1(\Gamma)$ is the completion $C^\infty(\Gamma)$.