Non-relativistic limit of complex scalar field 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 $\dot{\psi}$
 A: 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.
