When constructing the effective Lagrangian, we parametrize the Goldstone bosons (such as the pions $\pi_a$) by $U = \exp(i \pi_a \tau_a/2 f)$, where $\tau_a$ are the Pauli matrices. (See e.g. here). Under transformations by $SU(N_f)_L\times SU(N_f)_R$, this transforms as $U \rightarrow L U R^\dagger$. We then construct the most general Lagrangian using invariant traces of this., such as $Tr(\partial_\mu U \partial^\mu U^\dagger)$.
If we want to include other fields, background or dynamic, we add source terms to the underlying QCD Lagrangian, for example $v_\mu \bar q \gamma^\mu q$ for a vector current. We then demand that the theory is invariant under local $SU(N_f)_L\times SU(N_f)_R$, and enforce this by treating the source terms as gauge fields. This gives them transformation rules under $SU(N_f)_L\times SU(N_f)_R$, and we can add them to the effective, chiral Lagrangian, while still maintaining the symmetry.
To my question, when we have a dynamic photon field, the QCD lagrangian will have the vector current $v_\mu = e Q A_\mu$, where $Q $ is the quark charge matrix. As with other vector currents, for example a chemical potential, this is included in the effective theory by the covariant derivative, $\nabla U = \partial_\mu U - i [v_\mu, U]$. However, in addition, we also treat give the charge matrix transformation properties, $Q_R \rightarrow R Q_R R^\dagger$, and similar for $Q_L$. This leads to the term $Tr(QUQU^ \dagger)$. See here, or here.
Why do we give $Q$ these transformation properties? It seems unmotivated. All other transformation properties are derived by how they appear in the QCD Lagrangian, even the spurion field which gives the quarks their mass. The vector field already has the transformation law $v_\mu + a_\mu \rightarrow R(v_\mu + a_\mu + i \partial_\mu) R^\dagger$, ad similar for $L$. Does not the additional law for $Q$ destroy this?