I think you are mixing up two different things:
First, you can see QM as $0+1$ (one temporal dimension) QFT, in which the position operators (and their conjugate momenta) in the Heisenberg picture play the role of the fields (and their conjugate momenta) in QFT. You can check, for instance, that spatial rotational symmetry in the quantum mechanical theory is translated into an internal symmetry in QFT.
Secondly, you can take the "non-relativistic limit" (by the way, ugly name because Galilean relativity is as relativistic as Special relativity) of Klein-Gordon or Dirac theory to get "non-relativistic" Schrödinger QFT, where $\phi$ (in your notation) is a quantum field instead of a wave function. There is a chapter in Srednicki's book where this issue is raised in a simple and nice way. There, you can also read about spin-statistic theorem and the wave function of multi-particle states. Let me add some equations that hopefully clarify that (I'm using your notation and of course there may be wrong factors, units, etc.):
The quantum field is:
$$\phi \sim \int d^3p \, a_p e^{-i(p^2/(2m) \cdot t - p \cdot x)}$$
The Hamiltonian is:
$$H \sim i\int d^3x \left( \phi^{\dagger}\partial_t \phi - \frac{1}{2m}\partial _i \phi ^{\dagger} \partial ^i \phi \right) \, \sim \int d^3p \, \frac{p^2}{2m} \,a^{\dagger}_p a_p$$
The evolution of the quantum field is given by:
$$i\partial _t \phi \sim [\phi, H] \sim -\frac{\nabla ^2 \phi}{2m}$$
1-particle states are given by:
$$|1p\rangle \sim \int d^3p \, \tilde f(t,p) \, a^{\dagger}_p \, |0\rangle $$
(one can analogously define multi-particle states)
This state verifies the Schrödinger equation:
$$H \, |1p\rangle=i\partial _t \, |1p\rangle$$ iff
$$i\partial _t \, f(t,x) \sim -\frac{\nabla ^2 f(t,x)}{2m}$$
where $f(t,x)$ is the spatial Fourier transformed of $\tilde f (t,p)$.
$f(t,x)$ is a wave function, while $\phi (t, x)$ is a quantum field.
This is the free theory, one can add interaction in a similar way.
There's a major difference between Schrödinger/Dirac equations and Klein-Gordon one: the former are complex while the latter is real. But if you think of them a little, you'll also find a major similarity.
If you represent complex numbers of the form $a+ib$ with matrices of the form $\pmatrix{a&-b\\
b&a}$, then you can easily rewrite Schrödinger's equation like this (taking all dimensional constants equal to $1$):
$$\left\{\begin{align}
\dot R&=\hat H_RI-\hat H_I I\\
-\dot I&=\hat H_RR+\hat H_I R,
\end{align}\right.$$
where $R$ and $I$ are real and imaginary parts of the wavefunction $\psi=R+iI$, and Hamiltonian $\hat H=\hat H_R+i\hat H_I$.
Now Klein-Gordon equation can also be rewritten in this form:
$$\left\{\begin{align}
\dot\varphi&=A\\
\dot A&=\nabla^2\varphi-\mu^2\varphi.
\end{align}\right.$$
In both cases you have two simultaneous equations of the first order. In both cases you have to specify two initial conditions. For Schrödinger's equation they are real $R$ and imaginary $I$ parts of the wavefunction, and for Klein-Gordon equation they are $\varphi$ and $\dot\varphi$.
Best Answer
The Klein-Gordon equation does not replace the Schrödinger equation. The former is an equation for the field operators, the latter an equation governing the time evolution of the state.
See for example this, this or this.