What is the point of complex fields in classical field theory?

Two real scalar fields $\phi_1$ and $\phi_2$ satisfying an $SO(2)$ symmetry and one complex scalar field $\psi$ are equivalent. However, the latter is more convenient because the particles made by $\psi$ and $\psi^\dagger$ are each others' antiparticles. In the real case, the fields that have this property are $\phi_1 \pm i \phi_2$, so once you change basis from $\phi_1$ and $\phi_2$ to $\phi_1 \pm i \phi_2$ you've reinvented the complex scalar field.

This is explained nicely starting from p.53 in Sidney Coleman's QFT notes.

As you said, a complex quantity is not measurable in QM. And indeed $\psi$ is not an observable, which feels strange because quantum fields are often motivated at the start of a QFT course as local observables. Unfortunately this motivation isn't quite right, as we rarely measure quantum fields directly. For example, the number density, charge density, and current density for a charged complex scalar field are all field bilinears like $\psi^\dagger \psi$, and hence real.

In my viewpoint, the free complex field theory $$\tag 1 \mathcal{L}=\partial_\mu\phi\partial^\mu\phi^*-\frac{1}{2}m^2\phi\phi^*$$ is actually equivalent to the free double real field theory $$\tag 2 \mathcal{L}=\partial_\mu\phi_i\partial^\mu\phi_i-\frac{1}{2}m^2\phi_i\phi_i,$$ where $i=1,2$. To see this, simply write $\phi=\phi_1+i\phi_2$ and you can obtain Eq.(2) from Eq.(1). Eq.(1) is simply a compact form of Eq.(2). If you like you can define $\psi=\left(\phi_1,\phi_2\right)^T$ and have the form $$\tag 1 \mathcal{L}=\partial_\mu\psi^T\partial^\mu\psi-\frac{1}{2}m^2\psi^T\psi$$.

The double real field Lagrangian is, of course, Lorentz invariant. It also can represent charged particles since it also has $O(2)$ invariance and we can identify the corresponding Noether charge as electric charge.

BUT, there indeed a very crucial difference between the complex field and double real field. $\phi$ and $\phi^*$ ''have'' the opposite charge while $\phi_1$ and $\phi_2$ do not (please convince yourself by looking into $\phi=\rho e^{i\theta}$ and $\phi^*=\rho e^{-i\theta}$ and the transformation of O(2) is actually a rotation for the $\theta$). Since the observables are the charges, so $\phi$ and $\phi^*$ represent the particles with opposite charge while $\phi_1$ and $\phi_2$ can not.