The field and the wavefunction look similar, but they don't really have much to do with each other. The main point of the field is to group the creation and annihilation operators in a convenient way, which we can use to construct observables. As usual I will start with the free theory.
If we want to find a connection to non-relativistic QM, the field equation is not the way to go. Rather, we should look at the states and the Hamiltonian, which are the basic ingredients of the Schrödinger equation. Let's look at the Hamiltonian first. The usual procedure is to start with the Lagrangian for the free scalar field, pass to the Hamiltonian, write the field in terms of $a$ and $a^\dagger$, and plug that into $H$. I will assume you know all this (it's done in every chapter on second quantization in every QFT book), and just use the result:
$$H = \int \frac{d^3 p}{(2\pi)^3}\, \omega_p\, a^\dagger_p a_p$$
where $\omega_p = \sqrt{p^2+m^2}$. There's also a momentum operator $P_i$, which turns out to be
$$P_i = \int \frac{d^3 p}{(2\pi)^3}\, p_i\, a^\dagger_p a_p$$
Using the commutation relations it is straightforward to calculate the square of the momentum, which we will need later:
$$P^2 = P_i P_i = \int \frac{d^3 p}{(2\pi)^3}\, p^2\, a^\dagger_p a_p + \text{something}$$
where $\text{something}$ gives zero when applied to one particle states, because it has two annihilation operators next to each other.
Now let's see how to take the non-relativistic limit. We will assume that we are dealing only with one-particle states. (I don't know how much loss of generality this is; the free theory doesn't change particle number so it shouldn't a big deal, and also we usually assume a fixed number of particles in regular QM.) Let's say that in the Schrödinger picture we have a state that at some point is written as $|\psi\rangle = \int \frac{d^3 k}{(2\pi)^3} f(k) |k\rangle$, where $|k\rangle$ is a state with three-momentum $\mathbf{k}$. $f(k)$ should be nonzero only for $k \ll m$. Now look what happens if we apply the Hamiltonian. Since we only have low momentum, over the range of integration we can approximate $\omega_p$ as $m+p^2/2m$ and ignore the constant rest energy $m$.
$$H|\psi\rangle = \int \frac{d^3p}{(2\pi)^3} \frac{p^2}{2m} a^\dagger_p a_p \int \frac{d^3 k}{(2\pi)^3} f(k) |k\rangle \\
= \int \frac{d^3p}{(2\pi)^3} \frac{d^3 k}{(2\pi)^3} \frac{p^2}{2m} f(k) a^\dagger_p a_p |k\rangle \\
= \int \frac{d^3p}{(2\pi)^3} \frac{d^3 k}{(2\pi)^3} \frac{p^2}{2m} f(k) (2\pi)^3 \delta(p-k) |k\rangle \\
= \int \frac{d^3 k}{(2\pi)^3} \frac{k^2}{2m} f(k) |k\rangle = \frac{P^2}{2m} |\psi\rangle
$$
So if $|\psi\rangle$ is any one-particle state (which it is because the states of definite momentum form a basis), we have that $H|\psi\rangle = P^2/2m |\psi\rangle$. In other words, on the space of one-particle states, $H = P^2/2m$. The Schrödinger equation is still valid in QFT, so we can immediately write
$$\frac{P^2}{2m} |\psi\rangle = i \frac{d}{dt} |\psi\rangle$$
This is the Schrödinger equation for a free, non-relativistic particle. You will notice that I kept some concepts from QFT, particularly the creation and annihilation operators. You can do this no problem, but working with $a$ and $a^\dagger$ in QM is not particularly useful because they create and destroy particles, and we have assumed the energy is not high enough to do that.
Handling interactions is more complicated, and I fully admit I'm not sure how to include them here in a natural way. I think part of the issue is that interactions in QFT are quite limited in their form. We would have to start with the full QED Lagrangian, remove the $F_{\mu\nu}F^{\mu\nu}$ term since we aren't interested in the dynamics of the EM field itself, maybe set $A_i = 0$ if we don't care about magnetic fields, and see what happens to the Hamiltonian. Right now I'm not up to the task.
I hope I can convinced you that this newfangled formalism reduces to QM in a meaningful way. A noteworthy message is that the fields themselves don't carry a lot of physical meaning; they're just convenient tools to set up the states we want and calculate correlation functions. I learned this from reading Weinberg; if you're interested in these kinds of questions, I recommend you do so too after you've become more comfortable with QFT.
Best Answer
I have a slightly different perspective from the other two answers which provides a more elementary motivation. Suppose you know nothing about renormalizability or energy-momentum relations and all you know is that a Lagrangian density is a function of fields and their derivatives that transforms as a scalar under Poincaré transformations.
You can motivate the Klein-Gordon equation by asking what is the simplest Lagrangian you can write down for a scalar field that transforms as a scalar and provides a positive-definite Hamiltonian.
Since we're dealing with scalar fields any polynomial function of the fields $\phi$ will satisfy the correct Lorentz transformation property. So you could write down a term like $a\phi+b\phi^2$ with real constants $a$ and $b$. Now we also want to include derivatives $\partial_\mu\phi$. In order to satisfy the correct Lorentz transformation properties we need to contract this with a term $\partial^\mu\phi$.
So the simplest Lagrangian we can write down is $\mathcal{L}=c\partial_\mu\phi\partial^\mu\phi+a\phi+b\phi^2$ from which we obtain a Hamiltonian
$\mathcal{H}=\frac{\pi^2}{4c}+c\partial_i\phi\partial_i\phi-a\phi-b\phi^2$
The $a\phi$ term is not nice since it ruins the positive-definiteness of the Hamiltonian, so set $a=0$. A scalar field and the derivatives both have dimension of $[mass]^2$ and the Lagrangian density has dimension $[mass]^4$, so $c$ should be dimensionless and $b$ should be $b=-m^2$ where $m$ has units of mass and the minus sign is there to make the Hamiltonian positive-definite.
So we've reduced our Hamiltonian to
$\mathcal{H}=\frac{\pi^2}{4c}+c\partial_i\phi\partial_i\phi+m^2\phi^2$
Setting $c=1/2$ and rescaling $m^2\rightarrow m^2/2$ means the coeffecients of all terms are the same.
Hence the Lagrangian densiy is $\mathcal{L}=\frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\frac{1}{2}m^2\phi^2$
*Things you could try and argue against this being the simplest scalar field Lagrangian;
Of course, you get a simpler valid Lagrangian by setting $m=0$, but this isn't done as books want to show the energy-momentum relation in a general setting when you quantize the field. However, you can start with the massless case in 5 dimensions and perform dimensional reduction to obtain the massive case in 4 dimensions.