Thinking about spin triplet and singlet states in QFT

Actually, I don't completely understand your text of your question. In fact, it's not completely clear for me what you mean with " understand the spin state the pair of spin-1/2 particles by standard angular-momentum addition procedure:". However, in the following I assume that an "of" is missing if fitted as: "understand the spin state of the pair of spin-1/2 particles by ..." would make the text much clearer. You speak a lot of spinors (of different types), it would be much clearer if they were replaced by the word "spin-1/2 particles".

Under these interpretations I make a try to explain. Spin-singlet's and Spin-triplet states exist, for instance in the positronium, the bound state of an electron and a positron. If you consider the spin-singlet of the positronium which is called in this case Para-positronium, whereas the spin-triplet is called Ortho-positronium.(Both states are usually considered in relatistic quantum mechanics in particular in order to describe their decay into photons, a process which can only be treated in rel. QM. So a priori, one would describe both spin-1/2 particles, electron and the positron, with Dirac spinors).

However, both states can also be considered in the ground state like in s-state as in the H-atom. Its energy levels as virtually identical with those of a H-atom apart from mass differences of the constituents:

$E^H_n = -frac{R_E}{n^2}$ where $R_E= frac{1}{2}(m_e c^2) alpha^2$ is the Rydberg constant. For a positronium, however, the electron mass has to replaced by the reduced mass of the system $m_r = m_efrac{1}{1+m_e/m_e}= m_e/2$. (For the H-atom it is hardly necessary as $m_r approx m_e$ due to the large mass of the proton). Therefore (as an example) the lowest energy level is not -13.6eV, but -6.8eV as $E^{positronium}_n =-frac{R_E}{2n^2}$. And as it's well-known the energy levels of a H-atom (and as shown here for the positronium too) are result of non-relativistic quantum mechanics. As shown here the difference which appear are only due the different masses of the constituents and not due to spin states.

But apparently there are subtle differences in the energy levels due to relativistic or non-relativistic treatment, which can only be found by solving the Breit-equation or the Bethe-Salpeter-equation. These differences might not significantly related to the different spin states. In fact, this no longer belongs to the standard curriculum of physics study at the university, so I don't know these. But a good source for it is Landau/Lifschitz Volume 4, Relativitic quantum mechanics (admittedly rather old now). But, a priori, I don't expect any significant differences of singlet or triplet constitution of 2 spin-1/2 particles in the relativistic or non-relativistic case. Double quark systems can constitute spin singlet as well triplets even though these particles are usually described by the Dirac equation. The full mathematics of the transition from the relativistic case to the non-relativistic case as already mentioned, can be checked by studying the Breit equation. And don't expect deep insights at looking at the Breit equation. Actually, Landau/Lifschitz apply a rather pragmatical approach, there is no group-theoretical explanation at all how to deal with different spin representations. Briefly, singlet-spin and triplet-spin states of two 1/2 spin particles exist in non-rel. as well in rel. quantum physics, and a priori there is no significant difference.