The expression $$\langle x|H|\Psi(t)\rangle =H\langle x|\Psi(t)\rangle \ \ \ \ \ (\text{not true !!!})$$ You can insert an identity in between, $$\langle x|H|\Psi(t)\rangle = \int dx'\ \langle x|H|x'\rangle \langle x'|\Psi(t)\rangle$$ That's it, This is How far you can go without putting, $H=\frac{P^2}{2m}+V(X)$. $$ \int dx'\ \langle x|H|x'\rangle \langle x'|\Psi(t)\rangle=\frac{1}{2m}\int dx'\langle x|P^2|x'\rangle \psi(x',t)+\int dx'\langle x|V(X)|x'\rangle \psi(x',t)$$ Putting the matrix elements, leads to Shrodinger's equation.

The expression $$\langle x|H|\Psi(t)\rangle =H\langle x|\Psi(t)\rangle \ \ \ \ \ (\text{not true !!!})$$ You can insert an identity in between $$\langle x|H|\Psi(t)\rangle = \int dx'\ \langle x|H|x'\rangle \langle x'|\Psi(t)\rangle$$ That's it. This is how far you can go without putting $H=\frac{P^2}{2m}+V(X)$. $$ \int dx'\ \langle x|H|x'\rangle \langle x'|\Psi(t)\rangle=\frac{1}{2m}\int dx'\langle x|P^2|x'\rangle \psi(x',t)+\int dx'\langle x|V(X)|x'\rangle \psi(x',t)$$ Putting the matrix elements, leads to Shroedinger's equation.