We know photons have spin s=1. However, in Nuclear physics, the conservation of angular momentum in case of Gamma transitions is employed as follows: $$\vec J_i=\vec J_f+\vec L$$ where $J_i$ is the nuclear "spin" (total angular momentum spin+orbital contribution) of the parent nucleus, $J_i$ is the nuclear "spin" daughter nucleus and $\vec L$ is the orbital angular momentum of the photon. But we know that photons also carry spin angular momentum. Why we do not include the spin of the photon and write the conservation equation as $$\vec J_i=\vec J_f+\vec L+\vec S ?$$
Massless particles aren't characterized by spin because square of Pauli-Lubanski operator (which is Casimir operator of Poincare group) for them is equal to zero. They are characterized by helicity.
Now, let's back to conservation law. We don't include spin of photon in conservation law, because we can't introduce it. Instead of it we require that quantity $\mathbf L$ may have values $1, 2, ...$
In the reaction you write the photon takes away the angular momentum as its spin. Before the transition, the photon does not exist, so as to be included in the sum of quantum numbers. It appears at the transition as the carrier of the energy momentum and angular momentum .
This is clear in simple Feynman diagrams , for example
the decay of a sigma_0 to a photon and a lamda_0
the quantum numbers are balanced at the vertex. The photon and the lamda spins , momentum and energy must add up to the sigma_0 mass and spin.
In the complex environment of nuclei this still holds.