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.
First, observe that the Slater determinant you have written down is the linear combination of the singlet state and the z-spin-0 state of the triplet. Vice versa, you can produce the singlet and triplet states as linear combinations of Slater determinants.
Whether you prefer the Slater determinant or the singlet/triplet formalism for writing down your two-fermion states depends on your personal preference and on the specific application you have in mind.
Best Answer
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 $m$.