Photons have a spin of magnitude $1$. If we project the corresponding spin state vectors onto the momentum vector of the photon, we get what's called "helicity" with values $\pm1$ which are left and right circular polarization. Helicity 0 is forbidden for massless particles like photons. (You'll have to take it up with the physicists to really explain why. We'll just leave it at that.)
Since the process you describe involves two separate photon emissions, and each one changes the angular momentum of the atom by a magnitude of $1$, the photons that are emitted must also each carry a defined angular momentum to compensate the angular momentum change in the atom. The question is now: Which way are those angular momentum vectors (i.e., spin states) oriented?
Let's say that we place the calcium atom at the center of a Cartesian laboratory coordinate system. Let's further assume that the first photon is emitted along the $+z$ direction with a helicity of $+1$, i.e., the spin vector also points along the $+z$ axis. This would be right circularly polarized light. The atom, meanwhile, gains an angular momentum of opposite sense, described by a vector pointing along the $-z$ direction. Now, the second photon is emitted along the $-z$ axis, carrying away the angular momentum of the atom so that the latter returns to its $J=0$ state. This means that the second photon has its spin vector in the $-z$ direction; again, the helicity (projection of the spin onto its momentum) is $+1$, which is right circular polarization.
Now, let's talk about linear polarization some more, because it gets a little tricky.
A general photon whizzing though free space will have the same energy and momentum no matter whether it's in a $+1$ or $-1$ helicity state. Thus, we can put the photon into any superposition of the helicities and it will still have the same energy and momentum. (In a chemist's mind, think how combinations of complex-valued atomic orbital wavefunctions to the real-valued Cartesian ones still represent stationary solutions.)
This means that, in the general case, describing linear polarization as a superposition of circular polarization states is just as valid as the other way around.1 In the calcium experiment, we are perfectly able to describe the photons as being in a superposition state of horizontal and vertical linear polarization. But if we were to send the through linear polarizers to measure which polarization they have, each one would have a 50% chance for either result.
Note also that a single photon can actually have linear polarization if it is in a equally-weighted superposition state of $+1$ and $-1$ helicity. This does not mean that it has a helicity of $0$; instead, it means that when we measure the helicity, we have a 50% chance to get either result.2