If you'd like to see a small computation to show why microcausality is related to the vanishing of the commutator, here is a simple exercise that one can do.
Consider some operator $A(\vec{x},t)$ of which I want to measure the vacuum expectation value in some state $\psi$
$$
\mathcal{E}_A(\vec{x},t) := \langle \psi|A(\vec{x},t)|\psi\rangle\,.
$$
Now give a "kick" to the Hamiltonian at a certain time $t_0$ (let's assume $t_0 = 0$). By that I mean that we perturbe the Hamiltonian by some operator that is non zero only for $t > 0$. Namely
$$
H = H_0 + \theta(t)\, V(t)\,.
$$
How does the expectation value of $A$ change after this perturbation? It seems that the most convenient approach would be the interaction picture, so let's do that. Without reviewing the details, we define the state $|\psi\rangle$ and the operators $\mathcal{O}$ as the time evolution operator $\exp(i H_0 t)$ applied on the Schrödinger picture
$$
\psi_{\mathrm{int}}(t) = e^{i H_0 t} \psi_{\mathrm{S}}(t)\,,\qquad H_{\mathrm{int}}(t) = e^{i H_0 t}H e^{-i H_0 t}\,.
$$
The time evolution operator $U(t,t_0)$ must satisfy
$$
i \frac{\mathrm{d}}{\mathrm{d}t} \psi_{\mathrm{int}}(t) := i \frac{\mathrm{d}}{\mathrm{d}t} U(t,t_0) \psi_{\mathrm{int}}(t_0) = \theta(t) V(t)\,U(t,t_0) \psi_{\mathrm{int}}(t_0)\,,
$$
where the first equality is a definition of $U$ and the second its differential equation. To first order in the perturbation $V$ the solution is
$$
U(t,t_0) = \mathbb{1} - i \int_{t_0}^t\mathrm{d}t' \,V(t') + O(V^2)\,,\qquad \forall\;t_0 > 0\,.
$$
So far all standard. The expectation value can be then seen to transform as
$$
\begin{aligned}
\mathcal{E}_A(\vec{x},t) &= \langle\psi(t)|A(\vec{x},t)|\psi(t)\rangle \\&=
\langle \psi|U^\dagger(t,0)\, e^{i H_0 t} A(\vec{x},0)e^{-i H_0 t}\, U(t,0) |\psi\rangle \\&
\simeq \mathcal{E}_A(\vec{x},0) - i\int_0^t \mathrm dt'\langle \psi| A(\vec{x},t) V(t') - V(t') A(\vec{x},t) | \psi \rangle\,.
\end{aligned}
$$
Here I simply used all the definitions of the interaction picture and expanded to first order in $V$. Now let us make a physical assumption. This is similar to what one does in linear response theory. See the Kubo formula for instance.
The perturbation $V$ that I defined to be a "kick" happens not only at some specific time, but also at a specific location. Therefore it will modify the Hamiltonian as the integral of some local operator $B$. Namely
$$
V(t) = \int \mathrm{d}^{d-1} x B(\vec{x},t)\,.
$$
From this one has
$$
\mathcal{E}_A(\vec{x},t) - \mathcal{E}_A(\vec{x},0) = \int_0^t\mathrm{d}t'\int \mathrm{d}^{d-1} x'\,\langle\psi|\big[A(\vec{x},t)\,, B(\vec{x}{}',t')\big]|\psi\rangle\,.
$$
Here you see immediately that microcausality must imply that the correlator has to vanish outside the light cone. Suppose that $B$ creates a perturbation in some location in spacetime, it is impossible that $A$ knows about it if they are space-like separated. You would have to wait at least the time it takes for the light to get there in order to have a change in the expectation value. Therefore the only way to preserve causality is to require
$$
\big[A(\vec{x},t)\,, B(\vec{x}{}',t')\big] = 0\quad \mathrm{if}\; (x-x')^2 < 0\,.
$$
A simple contradiction that one might cook up is the following: tell a friend to make a perturbation to the Hamiltonian at time $t = 0$, or to not do it. Then you set yourself spacelike separated from your friend. If $A$ and $B$ do not commute you can infer whether your friend has decided to perturb be Hamiltonian or not by just measuring $\mathcal{E}_A$. And as you might know this leads to all sort of paradoxes in special relativity.
