We know that the many-electron wave function has to be antisymmetric with respect to the exchange of two electrons:
$\Psi (x_1,x_2,\dots,x_i,\dots,x_j,\dots)=-\Psi (x_1,x_2,\dots,x_j,\dots,x_i,\dots)$.
This symmetry is satisfied by Slater determinants a.k.a. electron configurations, so the wave function can be written in terms of them as
$|\Psi\rangle = \sum_I C_I |I\rangle$. Since determinants are orthonormal, $\langle I | J \rangle = \delta_{IJ}$, the wave function is normalized if $\sum_I C_I^2 =1$.
Each Slater determinant is defined by a set of orbitals that are occupied. The orbital basis can be chosen in many ways, but Hartree-Fock orbitals are a traditional choice. The Hartree-Fock determinant is the one where you occupy the orbitals with the lowest orbital energies; the orbitals themselves are defined by minimizing the expectation energy for the wave function.
One can also build more sophisticated wave functions on top of the Hartree-Fock reference as
$|\Psi\rangle = C_0 |0\rangle + \sum_S C_S |S\rangle + \sum_D C_D |D\rangle + \sum_T C_T |T\rangle + \dots $ where $|0\rangle$ is the Hartree-Fock determiant, and S, D, T denote singly, doubly, and triply excited determinants from the Hartree-Fock reference determinant. This approach is called configuration interaction theory. If you only include singles and doubles, you get configuration interaction singles and doubles (CISD); if you include all excitations, you have full configuration interaction (FCI). You can also only include excitations within a subset of the orbitals; this gives you the complete active space (CAS) model.
In cases where the wave function is single reference, i.e., $C_0 \approx 1 \iff C_0^2 \gg \sum_{I>0} C_I^2$ in the exact i.e. FCI wave function, the Hartree-Fock reference is qualitatively correct, and you can compute very accurate wave functions with e.g. the coupled cluster method.
In contrast, cases where $C_0$ is not large are known as multireference a.k.a. strongly correlated a.k.a. systems dominated by static correlation. Here, the electronic configurations become entangled either via small energy gaps or large exchange integrals.
A standard example is the Cr$_2$ molecule - a poster child for strongly correlated systems - where $C_0 \approx 10^{-4}$ for the Hartree-Fock wave function, so Hartree-Fock orbitals are spectacularly bad. Taking the orbitals from density functional calculations often lead to much better results; for Cr$_2$ IIRC one gets something like $C_0 \approx 0.6$, which is still very strongly correlated. Examining the nature of the correlations via a cluster decomposition reveals that the correlations in Cr$_2$ are highly non-trivial; see J. Chem. Phys. 147, 154105 (2017).
Note that instead of taking the orbitals from Hartree-Fock or density functional calculations, the "correct" way is to relax the orbitals for the used many-electron wave function, instead. For CAS this gives the CAS self-consistent field method, CASSCF, which can be contrasted to "single-point" CAS-CI calculations at fixed orbitals.
Let's say we have a molecule A which has static correlation effects and it has two dominant configurations. So, would that mean that some A molecules are in one electronic configuration and the other A molecules are in another configuration at one moment? Or does it mean that each A molecule exists as a mix of the two configurations?
This means that the many-electron wave function for molecule A will be a superposition of the two configurations: $|\Psi\rangle = C_1 |1\rangle + C_2 |2\rangle$ with $|C_1| \gg 0$ and $|C_2| \gg 0$. You may even have $|C_1| = 2^{-1/2} = |C_2|$.
If a molecule is a mix of two configurations, then how do we calculate its spin multiplicity (I think CASSCF calculations in Orca print out spin)?
The spin multiplicity is actually trivial: the Hamiltonian commutes with $\hat{S}_z$, so your two configurations 1 and 2 have to have the same multiplicity i.e. the same number of alpha electrons and beta electrons in order to mix in the wave function.
The expectation value of the total spin, $\langle \hat{S}^2 \rangle$ is more complicated. A proper eigenfunction satisfies $\hat{S}^2 \Psi_S = S(S+1) \Psi_S$, and it also has $2S+1$ possible z projections: $\hat{S}_z \Psi_S = S_z \Psi_S$ with $S_z \in [-S, -S+1, ..., S-1, S]$. Since you know the many-electron wave function, you can just go ahead and compute the value of $\langle \hat{S}^2 \rangle$.
It is important to note that states with different $\langle \hat{S}^2 \rangle$ can appear in your calculation for a given $S_z$: the state you get may not be the one you wanted! For instance, if you're looking for a singlet, you might get a quintet wave function instead!
How would we calculate ionization energies, if the electron is present in different orbitals simultaneosuly?
Easy, just the way you would do it in any other method. You calculate the neutral molecule and its cation, and substract the total energies. For electron affinities you'd do the anion and the neutral.
Are orbital occupation numbers fractional?
This is a bit tricky, since the one-electron density matrix may not even be diagonal! If you diagonalize the matrix, you get natural orbitals, and they do have fractional occupation numbers. But, orbitals don't really have much meaning when you go beyond the self-consistent field level of theory.
And finally, what does Frank Jensen mean when he says electrons avoid each other on a "permanent" basis?
I'm not sure what this is referring to; however, once you allow for multiple configurations in the wave function, you are also allowing the electrons more variational freedom.
If you look at Hartree-Fock, the probability to find an electron at ${\bf r}$ and an electron of opposite spin at ${\bf r}'$ is independent of the distance $|{\bf r}-{\bf r}'|$. This means it is possible to find two electrons arbitrarily close together, even though the Coulomb interaction blows up. When you include more configurations, this probability decays when $|{\bf r}-{\bf r}'|$ is small, that is, the electrons avoid each other. A similar argument also holds for same-spin electrons; however, the error in Hartree-Fock is much smaller in this case because of the Pauli exclusion principle, so same-spin correlation energy is very small compared to the opposite-spin correlation energy.
