There are lots of questions asked here, but I'll attempt to answer some of these...
Oxygen is found in the triplet state because the triplet state is most stable. This is a complex function of the properties of the atoms (e.g. charge and separation between atoms) and the electrons (e.g. number of electrons present, possible combinations of orbitals). The molecular orbitals given on the wiki page show three different states:
$^{3}\Sigma^{-}_{g}$: The ground triplet state
$^{1}\Delta_{g}$: The ground singlet state
$^{1}\Sigma^{+}_{g}$: An excited singlet state
These symbols are explained well at Wikipedia. In short, a triplet state has two electrons with parallel spins, for example the two red arrows pointing 'up' in the $^{3}\Sigma^{-}_{g}$ MO diagram.
If we investigate the energy of an oxygen molecule as a function of the distance between the oxygen atoms, we can uncover which of these states are most stable. An example of this is given here. It is evident that the $^{3}\Sigma^{-}_{g}$ state is lower than all other states, thus we expect triplet oxygen to be the most stable state.
More complicated electronic configurations do exist, an example of these is given by the $^{1}\Sigma^{+}_{g}$ state. We can see here that it is even less stable than ground singlet ($^{1}\Delta_{g}$) oxygen. These variations in energy states arise because 'putting' electrons into different orbitals with different spins give varied 'goodness of overlap'.
Thermodynamic properties would be expected to be subtly different between electronic states. This is because the molecule would have different vibrational states which would 'translate' into different thermodynamics through statistical mechanics. I cannot find a source which directly compares such properties.
The reaction features are, of course, very different. For example, triplet oxygen will happily dissolve in water, but singlet will react with it. Singlet oxygen is often used when one wants to 'attack' double bonds. An explanation of why this is the case is very complex - but in short, is due to the quantum state of oxygen compared with the states of most molecules it attempts to react with.
The spin part of the quantum state of any system consisting of two spin-$1/2$ particles (including a Deuterium nucleus) can be described as a general linear combination of the singlet and triplet states.
The the symbolic manipulation $2\otimes 2 = 3\oplus 1$ is telling you that the Hilbert space of the system of two spin-$1/2$ particles, which is simply the tensor product of the spin-$1/2$ Hilbert space with itself, admits an orthonormal basis for which $3$ of the states in the basis, the so-called triplet states, have total spin quantum number $s=1$, while one of the states in the basis, the so-called singlet state, has total spin quantum number $s=0$.
More explicitly, if we denote $|s, m\rangle$ as a state with
\begin{align}
S^2|s,m\rangle = \hbar^2 s(s+1) |s,m\rangle, \qquad S_z|s,m\rangle = \hbar m|s,m\rangle
\end{align}
where $S^2$ is the total spin squared operator, and $S_z$ is the $z$-component of the total spin operator, then the triplet states are
\begin{align}
|1,1\rangle, \qquad |1,0\rangle, \qquad |1, -1\rangle,
\end{align}
and the singlet state is
\begin{align}
|0,0\rangle,
\end{align}
and together they form a basis for the spin Hilbert space of the two-spin-$1/2$ system.
As can be seen explicitly in the notation, the triplet states are distinguished by the value of their "magnetic" quantum number $m$.
Best Answer
I am not sure that I get your question right, but let me try to answer according to my understanding.
The spin part of the two electron wave function of the singlet state $|0,0\rangle=|l=0,m=0\rangle$ is
$|0,0\rangle=(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle)/\sqrt{2}$
The three triplet states look like this:
$|1,0\rangle=(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle)/\sqrt{2}$
$|1,1\rangle=|\uparrow\uparrow\rangle$
$|1,-1\rangle=|\downarrow\downarrow\rangle$
If you are unfamiliar with the notation, $\uparrow$ denotes spin $+1/2$ and $\downarrow$ denotes spin $-1/2$; the numbers on the left denote: the total angular momentum $l$ of both electrons and its $z$ component $m$.
If the electrons are in the same energy state, they have to have different spins, that is, the state can be either $|0,0\rangle$ or $|1,0\rangle$, so in principle singlet and triplet are possible. If one of the electrons is excited, any of the four states is possible, since the spins don't have to be different any more. Therefore, if you know that both electrons have spin up or both have spin down, you can be sure, that it is a triplet state. If one spin is up and the other down, you cannot tell from the spin configuration whether it is a singlet or a triplet, since again both $|0,0\rangle$ or $|1,0\rangle$ are possible.