There are several misconceptions in this post. Let me address them one by one.
In non-relativistic quantum mechanics, causality is violated by saying that the amplitude of propagation of a particle $$A=\langle \textbf{x}|\exp{\Big(\frac{-i\textbf{p}^2t}{2m\hslash}}\Big)|\textbf{x}_0\rangle$$ between any two points $\textbf{x},\textbf{x}_0$ is non-zero for any time $t$, however small.
This is wrong on several different levels. First of all, in non-relativistic mechanics a process is acausal if and only if its amplitude is non-zero for $t<0$. Whether it vanishes or not for $t>0$ is irrelevant. Indeed, if $c\equiv \infty$ perturbations can propagate infinitely fast (cf. the heat equation), and this is not considered a violation of causality. The latter is violated if and only if effects are observed before their cause, i.e., if the amplitude is non-zero for negative times.
Second of all, your expression for $A$ is not a propagation amplitude; you forgot the step function (cf. wikipedia)
$$
A=\langle \textbf{x}|\exp{\Big(\frac{-i\textbf{p}^2t}{2m\hslash}}\Big)|\textbf{x}_0\rangle\Theta(t)
$$
Now you can see that $A$ vanishes for $t<0$, so the amplitude is causal, as required. So far so good.
But this is also true in quantum field theory. The amplitude of propagation of a particle from a spacetime point $x$ to another spacetime point $y$, given by $$A(x,y)=\langle 0|\phi(x)\phi(y)|0\rangle\neq 0$$ even for space-like separations.
No, this is not correct. The object $\langle 0|\phi(x)\phi(y)|0\rangle$ has nothing to do with a propagation amplitude, because the object $\phi(y)|0\rangle$ does not represent a particle localised at $y$. You already know that. A (non-) vanishing $A(x,y)$ has nothing to do with causality.
Therefore, isn't by the previous argument, here too, causality violated?
No, because the first $A$ represents a completely different object from the second $A$. Use the same letter if you want, but you are dealing with two different objects.
This is "resolved'' in QFT by saying that we should not ask whether particles can propagate over spacelike intervals, but whether a measurement at $x$ can affect a measurement $y$ if their separation is spacelike, and when one computes $[\phi(x),\phi(y)]$ for $(x-y)^2<0$, it turns out that the commutator vanishes. So causality is preserved.
This is somewhat correct, but far from being the whole story. Let me quote a paragraph from Weinberg's QFT (page 198):
The condition $(5.1.32)$ is often described as a causality condition, because if $x-y$ is space-like then no signal can reach $y$ from $x$, so that a measurement of $\psi_\ell$ at a point $x$ should not be able to interfere with a measurement of $\psi_{\ell'}$ or $\psi^\dagger_{\ell'}$ at point $y$. Such a consideration of causality are plausible for the electromagnetic field, any of whose components may be measured at a given spacetime point, as shown in a classic paper of Bohr and Rosenfeld. However, we will be dealing here with fields like the Dirac field of the electron that do not see in any sense measurable. The point of view taken here is that Eq.$(5.1.32)$ is needed for the Lorentz invariance of the $S$-matrix, without any ancillary assumptions about measurability or causality.
where $(5.1.32)\equiv [\psi_\ell(x),\psi_{\ell'}(y)]=[\psi_\ell(x),\psi^\dagger_{\ell'}(y)]=0$ for $x-y$ space-like.
So my question is why do we take two different approaches to the "meaning of causality" in quantum mechanics and quantum field theory?
We do not. Causality is just the statement that the effect comes after the cause. In non-relativistic mechanics every interval is time-like so you only have to worry about the step function. In relativistic mechanics, the propagation velocity of everything is finite so you have to worry about space-like intervals as well. But that's it.