In page 42 of David Tong's lectures on Quantum Field Theory, he says that one can also derive the Schrödinger Lagrangian by taking the non-relativistic limit of the (complex?) scalar field Lagrangian. And for that he uses the condition $\partial_{t} \Psi \ll m \Psi$, which in fact I suppose he means $|\partial_{t} \tilde{\Psi}| \ll |m \tilde{\Psi}|$, otherwise I don't get it. In any case, starting with the Lagrangian:
$$\mathcal{L}=\partial^{\mu}\tilde{\psi} \partial_{\mu} \tilde{\psi}^{*} -m^{2}\tilde{\psi}\tilde{\psi}^{*}$$
Using the inequation I think it's correct, I can only get to:
$$\mathcal{L}=-\nabla\tilde{\psi} \nabla \tilde{\psi}^{*} -m^{2}\tilde{\psi}\tilde{\psi}^{*}$$
And from that I've tried relating $\tilde{\psi}$ or $\psi$ (as we can write the above Lagrangian with both, as it's invariant under multiplying by a pure phase), to $\dot{\psi}$
Best Answer
You cannot derive it "directly" from a Klein-Gordon equation, or from a Klein-Gordon Lagrangian.
Starting from a Klein-Gordon equation for $\psi$, and defining $\psi(\vec x,t) = e^{-imt} \tilde \psi(\vec x,t)$ ($2.103$), you get a new equation for $\tilde \psi$, which is not a Klein-Gordon equation :
$$ \ddot {\tilde \psi} - 2 im \dot {\tilde \psi} - \nabla^2\tilde \psi = 0 \tag{2.104}$$
By Fourier transform , this is equivalent to the condition :
$$ (E'^2 + 2mE'-\vec p^2) = 0 \tag{1}$$
What does that mean?
We begin with a Klein-Gordon equation for $\psi$, which, by Fourier Transform, is equivalent to the condition $$ (E^2 -\vec p^2 - m^2) = 0 \tag{2}$$
Now, the transformation $\psi(\vec x,t) = e^{-imt} \tilde \psi(\vec x,t)$ $(2.103)$, gives the link between $E'$ and $E$, this is $E' = E - m$, this is a shift in the definition of the energy.
So, from $(2)$, we have simply : $((E'+m)^2-\vec p^2 - m^2)=0$, which is just the condition $(1)$
Now, if we suppose $|\vec p| \ll m$, this means $|E-m| \ll m $ (with $E \sim m)$, that is $E'\ll m$, so $E'^2 \ll mE'$.
Turning back to the equation $(2.104)$, which is not a Klein-Gordon equation, we see, by Fourier Transform, that we can neglect the first term relatively to the second term, and finally, you get :
$$ i \dot {\tilde \psi}= - \frac{1}{2m}\nabla^2\tilde \psi = 0 \tag{2.105}$$
About lagrangians, you will have, I think, a problem, if you want to define a lagrangian giving $(2.104)$, with only a real scalar field $\tilde \psi$, because of the $\dot {\tilde \psi}$ term.
So, you have to consider a complex scalar field, and in the Lagrangian, you will have terms like $ \dot{ \bar {\tilde \psi}} \dot{ {\tilde \psi}}$ and $im \bar {\tilde \psi}\dot{ {\tilde \psi}}, im \dot{\bar {\tilde \psi}}{ {\tilde \psi}}$, and the first term, in the approximation we discussed, is negligible relatively to the other terms.