Hmmm, an old question without a satisfactory answer. I'll have a go.
The spins of the two $B$ may combine as
\begin{align}
\text{singlet}\quad|s_1s_2\rangle &= \frac{\left|\uparrow\downarrow\right> - \left|\uparrow\downarrow\right\rangle}{\sqrt2},
&
\text{or triplet}\quad|s_1s_2\rangle &= \frac{\left|\uparrow\downarrow\right> + \left|\uparrow\downarrow\right\rangle}{\sqrt2}.
\end{align}
Since the two $B$ have spin-1/2, they obey Fermi-Dirac statistics and their total wavefunction must be antisymmetric under exchange. Therefore the antisymmetric spin singlet can only be paired with $L=0,2,\cdots$, which have even parity; the symmetric spin triplet must be paired with $L=1,3,\cdots$, so that the parity of the orbital angular momentum wavefunction makes the entire wavefunction change sign if the two $B$ are interchanged.
The intrinsic parity of $B$ doesn't contribute to the overall parity of the final state, since there are two of them; if the $B$ have negative parity, the overall intrinsic parity of the pair is still positive. Thus the allowed final states are
\begin{align}
\text{positive parity}&: & \text{spin singlet (antisymmetric)} && L&=0 \\
\text{negative parity}&: & \text{spin triplet (symmetric)} && L&=1
\end{align}
I've omitted the spin singlet combined with $L=2$, since we do not have that much angular momentum. Likewise there is no combination of the spin triplet and $L=3$ which can be produced from a $J=2$ initial state.
If $A$ has definite parity and parity is conserved in the decay, only one of these possibilities is allowed.
In fact, the way that angular momentum combines (with the possibility of the spin triplet plus the $L=1$ wavefunction adding to a $J^P=0^-$ final state) means that the parity has more to do with the allowed final spin state than $A$'s spin does:
$$
\begin{array}{r|cc}
& \text{spin 0} & \text{spin 1} \\
\hline \text{parity}+ & \text{decay to singlet},0^+ & \text{decay forbidden} \\
\text{parity}- & \text{decay to triplet},0^- & \text{decay to triplet},1^-
\end{array}
$$
"Spin parity" isn't a thing. It's saying the xi baryon has spin $\frac{1}{2}$ and positive parity; they're separate properties whose names tend to be run together for some reason.
As for why we use the word spin even though some of the angular momentum may be orbital: it allows you to imagine the $\Xi^-$ as an elementary particle which has the same amount of angular momentum as a spin-$\frac{1}{2}$ fundamental particle, namely $\frac{\hbar}{2}$. When you're treating the $\Xi^-$ as an elementary particle, you don't care whether part of its angular momentum comes from the orbital angular momentum of even more fundamental constituents. In a sense, all of quantum field theory is based on the idea that we can ignore small-scale structure in this way, under appropriate conditions.
Best Answer
The notation always refers to the total angular momentum. A source of confusion may be that in nuclear physics, we talk about the "spin" of a nucleus as a whole, even though the spin is only partly due to the intrinsic spin 1/2 of the neutrons and protons.