The issue is discussed in some detail in [1] , referring to perovskites as an example , applying second-order perturbation to the Hamiltonian
$$\mathcal{H}=\epsilon_d\sum_{i,\sigma}d^\dagger_{i,\sigma}d_{i,\sigma}+\epsilon_p\sum_{j,\sigma}p^\dagger_{j,\sigma}p_{j,\sigma}+\sum_{ij,\sigma\sigma'}\left(t_{\left( pd \right)ij}d^\dagger_{i,\sigma}p_{j,\sigma'} + h.c.\right) \\ +U_{dd}\sum_i n_{di\uparrow}n_{di\downarrow}+U_{pp}\sum_j n_{pj\uparrow}n_{pj\downarrow}+U_{pd}\sum_{ij,\sigma\sigma'} n_{di \sigma}n_{pj \sigma'}$$
where $d$ refers to $d$ orbital holes in the metalllic ions and $p$ to $p$ oxygen holes. In perovskites the transition metal ions are separated by oxygen ions, so that there is negligible direct hopping between metal ions. Instead there is a significant overlap of $d$ orbitals of transition metals with $2p$ oxygen orbitals, so that hopping of $d$ holes from one metal ion to the other goes over the intermediate oxygens (ligands). Since oxygens are the only hopping channel, one may factor them out, recovering the standard Hubbard model. However tracking them through the $t_{pd}p^{\dagger}d$ term
one sees that there are two hopping paths. The first one is $d^n\left(p^6\right)d^n \rightarrow d^{n+1}\left(p^6\right)d^{(n-1)}$, costing energy $U_{dd}$, where a $d$ hole hops from one ion to another and back via intermediate oxygen $p$ levels. The corresponding Heisenberg exchange constant is $$
J_1=\frac{2t_{pd}^4}{\Delta^2 U_{dd}}.
$$ In the second case after first transferring a $d$ hole to the oxygen from a $d$ ion, $d^n p^6 \rightarrow d^{n+1}p^5$, one transfers to the same oxygen another $d$ hole from another ion, so that the intermediate energy is different from the first case, since we have two holes on the same oxygen. The corresponding energy is $\Delta=\tilde{\epsilon_d}-\tilde{\epsilon_p}$ i.e. the energy difference between oxygen and
transition metal hole levels, or, more precisely taking into account the hole repulsion on the oxygen ions, $\Delta=\tilde{\epsilon_d}-\tilde{\epsilon_p}+U_{pp}$. This yields the exchange constant$$
J_2=\frac{4t_{pd}^4}{\Delta^2 \left(2\Delta + U_{pp} \right)}.
$$
where the factor is 4 instead of 2 because there are twice as many hopping routes in the second case.
The total interaction is
$$J_{tot}=J_1+J_2 =\frac{2 t^4_{pd}}{\Delta^2}\left(\frac{1}{U_{dd}} + \frac{2}{2\Delta+U_{pp}}\right). $$
Depending on whether the first or the second term in $J_{tot}$ is prevalent, one gets two different situations.
In both cases, if $t_{pd} << \left(U_{dd}, \Delta \right)$, the ground state is a Mott insulator, but there is a difference in the lowest charge-carrying excitations. In the first case (pure Mott) the excitation creates an extra $d$ electron and an extra $d$ hole, leaving the oxygen unchanged. In the second case (
$\Delta < U_{dd}$, ZSA charge-transfer, prevalent in oxides of heavier transition metals such as Cu), we get a doubly occupied $d$ state and an oxygen hole (cf. Pieter's answer above). Now, if one hole-dopes the system, the extra holes would go in the MH case to the metal d orbitals, in the ZSA case they would go to the oxygens. In concrete situations this difference may be significantly affect a.o.t. the distribution of electron and spin density.
Besides, materials on the left of the diagram above would have oxygen holes as charge carriers and "the properties in this regime may be quite nontrivial".
Seems enough to care.
[1] 12.10 Charge-transfer insulators in Khomskii , Basic Aspects of the Quantum Theory of Solids Order and Elementary Excitations. 2010
