Energy-Momentum tensor in the non-relativistic limit of Klein-Gordon Field Assume we have a real Klein Gordon field $\phi(x,y,z,t)$, and we do the non-relativistic expansion of it in terms of a complex field $\psi(x,y,z,t)$
$$\phi=\frac{1}{\sqrt{2m}}(\psi e^{-imt}+\psi^* e^{imt}).\tag{1}$$
Notice that in the non-relativistic limit, $\psi$ obeys the Schrodinger equation
$$i\dot{\psi}=-\frac{\nabla^2\psi}{2m}.\tag{2}$$
We have the energy momentum tensor of the Klein Gordon field
$$T^{00}=\frac{1}{2}\dot{\phi}^2+\frac{1}{2}(\nabla \phi)^2+\frac{1}{2}m^2\phi^2.\tag{3}$$
If we substitute $\phi$ in terms of $\psi$ and neglect all terms that are highly oscillating (i.e. containing factor of $e^{\pm 2imt}$ after the substitution ), we have the following expression of the $T^{00}$.
$$T^{00}=\frac{1}{4m}(2\dot{\psi}\dot{\psi^*}+2im\psi^*\dot{\psi}-2im\psi\dot{\psi^*}+4m^2|\psi|^2+2(\nabla \psi)\cdot(\nabla \psi^*)).\tag{4}$$
Then the energy of the Klein Gordon field should be given by $\int d^3x T^{00}$. Then the $4m^2|\psi|^2 $ term will integrate to give m, which is the rest energy of the particle
The $2\dot{\psi}\dot{\psi^*}$ term can be neglected, as it is suppressed by the non-relativistic factor compared with other term, for example
$$(\nabla \psi)\cdot(\nabla \psi^*)\sim p^2\psi^2,\tag{5}$$
$$\dot{\psi}\dot{\psi^*}\sim \frac{p^4}{m^2}\psi^2,\tag{6}$$
Where $p$ is approximately the momentum of the $\psi$ field, and $\frac{p}{m}\sim \frac{v}{c}\ll 1$ according to non-relativistic assumption.
However, it seems to me that the kinetic energy term is with the wrong coefficient, because
$$\frac{i\psi^*\dot{\psi}-i\psi\dot{\psi^*}}{2}+\frac{1}{2m}(\nabla \psi)\cdot (\nabla \psi^*)=\psi^*(-\frac{\nabla^2}{m}\psi).\tag{7}$$
The equation makes use of the Schrodinger equation and is up to total spatial derivative, which plays no role in integration
Then, it seems to me that
$$\int d^3x T^{00}=m+\langle\psi|-\frac{\nabla^2}{m}|\psi\rangle=rest \quad energy+ 2\times kinetic\quad energy.\tag{8}$$
That is, the coefficient of the kinetic energy is different from the naive  expectation $E_{rel}=E_{rest}+E_{kin}$ by 2.
Can anyone tell me what is happening here? Why is it off by a factor of 2? I also have the question: When we talk about energy, which energy are we talking about? Which energy is observable or physical? For example, the field may represents some axion cloud that evolves around a gravitational potential. And the gravitational potential provider (some star, for example) may exchange the energy with the axion cloud. Which energy should I use when we discuss this "backreaction" problem? The energy of the $\phi$ field or the energy of the $\psi$ field?
 A: *

*We can ignore the cross-terms in the Klein-Gordon (KG) action from OP's expansion
$$\Phi(\vec{x},t)~=~\frac{\hbar}{\sqrt{2m}}\left(\exp\left(-\frac{imc^2t}{\hbar}\right)\psi(\vec{x},t)
+\exp\left(\frac{imc^2t}{\hbar}\right)\psi(\vec{x},t)^{\ast}\right)\tag{1}$$
in the limit $c\to\infty$ due to Riemann-Lebesgue lemma. In order words, the particle-antiparticle  interaction terms disappear, so we might as well focus on the  particle sector alone
$$\Phi(\vec{x},t)~=~\frac{\hbar}{\sqrt{2m}}\exp\left(-\frac{imc^2t}{\hbar}\right)\psi(\vec{x},t).\tag{1'}$$
(The anti-particle sector is similar.)


*We will follow standard conventions: For a real (complex) KG field $\Phi$, the action terms have (have not) a $1/2$ symmetry factor, respectively. Eq. $(1)$ is a real KG field, while eq. $(1')$ is a complex KG field.


*It is straightforward to derive the corresponding Schrödinger Lagrangian density
$$\begin{align}
{\cal L}_{KG}~=~&\left|\frac{1}{c}\partial_t \Phi\right|^2 - |\nabla \Phi|^2 -|\frac{mc}{\hbar}\Phi|^2\cr \stackrel{(1')}{\longrightarrow}&\underbrace{\frac{i\hbar}{2}\left(\psi^{\ast}\partial_t\psi-\partial_t \psi^{\ast}\psi \right)}_{\text{symplectic terms}} 
-\underbrace{\frac{\hbar^2}{2m}|\nabla \psi|^2}_{\text{non-rel. Hamiltonian density}}\cr
&\quad {\rm for }\quad c~\to~\infty,\end{align}\tag{A}$$
cf. e.g. my Phys.SE answer here.


*The relativistic KG energy density is dominated by the rest energy
$$\begin{align}
{\cal T}^{00}_{KG}~=~&\left|\frac{1}{c}\partial_t \Phi\right|^2 + |\nabla \Phi|^2 +|\frac{mc}{\hbar}\Phi|^2\cr \stackrel{(1')}{\longrightarrow}&\frac{i\hbar}{2}\left(\psi^{\ast}\partial_t\psi-\partial_t \psi^{\ast}\psi \right) 
+\frac{\hbar^2}{2m}|\nabla \psi|^2
+\underbrace{mc^2|\psi|^2}_{\text{rest energy density}}\cr
\stackrel{\begin{array}{c}\text{TDSE}\cr\text{IBP}\end{array}}{\sim}& \frac{\hbar^2}{m}|\nabla \psi|^2
+\underbrace{mc^2|\psi|^2}_{\text{rest energy density}}\cr
&\quad {\rm for }\quad c~\to~\infty,\end{align}\tag{B}$$
and hence less useful in the non-relativistic limit.


*If we include the anti-particle sector and recall the normalization convention from section 2, then the RHS of eqs. (A) & (B) remain the same. Therefore eq. (B) agrees with OP's result (8).


*OP's issue is related to that the Legendre transformation from the Lagrangian to the Hamiltonian formulation grows singular in the limit $c\to\infty$, see e.g. this Phys.SE post. It is important to recognize that the symplectic terms are not part of the non-relativistic Hamiltonian in eq. (A); compare with OP's eq. (7).


*Whereas there is a smooth transition between the relativistic and non-relativistic actions, note that the non-relativistic notion of energy (which excludes rest energy) is a departure from the relativistic notion of energy (which includes rest energy). This feature is similar to what happens for the point particle, cf. e.g. my Phys.SE answer here.
A: You are mixing the real Klein-Gordon field, which describes a neutral particle with the complex Schrödinger field. That is not consistent.
The correct approach starts with the complex Klein-Gordon field. Then write $$\phi = e^{imt} \psi \,.$$ This gives $$\dot \phi = im e^{imt} \psi + e^{imt} \dot \psi \,.$$ The non-relativistic result is obtained by neglecting terms that are nonlinear in $\dot \phi$, which happens in the energy-momentum tensor, or contain the second time derivative, which happens in the equation of motion.
An approach that does not work is to take the non-relativistic limit of the Lagrangian and then derive the equation of motion in terms of $\psi$.
