For single particles, S orbitals have angular momentum quantum number $\ell = 0$, and P orbitals have $\ell = 1$. The magnetic quantum number $m$ runs from $-\ell$ to $+\ell$ in integer steps, so S orbitals can only have one value of $m=0$ (hence a singlet state), and P orbitals can have $m=-1,0,1$ (hence a triplet of states). Is this what you are asking?
Edit
What I think your instructor meant is along the following lines. For a two-electron state, the combined wavefunction of the two electron system needs to be antisymmetric under exchange of electrons. The total wavefunction is the tensor product of the spin and spatial parts, so if one is symmetric, the other needs to be antisymmetric.
Note that the singlet $|\downarrow \uparrow\rangle - |\uparrow \downarrow\rangle$, with $s=0$, is antisymmetric under particle exchange, so the wavefunction of the electrons needs to be symmetric. So we can arrange $\Psi(x_1, x_2)$ for the spatial part of the wavefunction to be symmetric. The theory of addition of angular momentum tells us that states with even angular momentum are symmetric under exchange, so in this case $\ell = \ell_1 + \ell_2 = 0, 2, \dots$ is allowed. For the S state ($\ell = 0$), we can just write $\Psi(x_1, x_2) = \psi_s(x_1) \psi_s(x_2)$ where $\psi_s$ is a single-particle wave-function with $\ell_{1,2} = 0$.
For the triplet states, with $s=1$, each spin state is symmetric under particle exchange, so we need to arrange a spatial wavefunction that is antisymmetric. Allowable spatial wavefunctions have $\ell = 1, 3,\dots$. The simplest is $\ell=1$ which is a symmetrized linear combination of $\psi_s$ and $\psi_p$. The exact form of the wavefunction depends on $m_\ell$.
Note though that we are doing all of this from the point of view of the spins. So that while the $S$ state has both particles individually in an $S$ state, the two-particle $P$ state has individual particles in a superposition of $S$ and $P$ states.
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
If we consider two spin $1/2$ particles with spin up and spin down states, then there are four possibilities in total: \begin{equation} \begin{array}{cccc} |++\rangle \; ,& |+-\rangle \; ,& |-+\rangle \; ,& |--\rangle \end{array} \end{equation} where this notation means for instance: \begin{equation} |+-\rangle = |s_1=1/2,m_1=1/2;s_2=1/2,m_2=-1/2\rangle \end{equation} We will suppose that the system consisting of both particle has zero orbital angular momentum and let $S_z$ denote that operator acting on the system. Then it is very easy to evaluate the following eigenvalue equations: \begin{align} S_z|++\rangle &=\hbar|++\rangle \\ S_z|+-\rangle &=0|+-\rangle\\ S_z|-+\rangle &=0|-+\rangle \\ S_z|--\rangle &=-\hbar|--\rangle \\ \end{align} and so $S_z$ can be written as: \begin{equation} S_z = \hbar \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} \end{equation} Now, if $s=1$ (remember $s$ denotes the quantum number for the total system), then the three states of the system are: \begin{align} |s=1,m=1\rangle & = |++\rangle \\ |s=1,m=0\rangle & = \sqrt{\frac{1}{2}}(|+-\rangle+|-+\rangle) \\ |s=1,m=-1\rangle & = |--\rangle \end{align} with eigenvalues: \begin{align} S_z |1,1\rangle & = \hbar |1,1\rangle \\ S_z |1,0\rangle & = 0|1,0\rangle \\ S_z |1,-1\rangle & = -\hbar |1,-1\rangle \end{align} where I have switched to a more compact notation: \begin{equation} |s=1,m=1\rangle \equiv |1,1\rangle \end{equation} However, we can also get an eigenstate with $s=0$: \begin{equation} |0,0\rangle = \sqrt{\frac{1}{2}}(|+-\rangle-|-+\rangle) \end{equation} with eigenvalue: \begin{equation} S_z |0,0\rangle = 0|0,0\rangle \end{equation} Now, we can also verify that: \begin{equation} S^2 \equiv S_x^2 + S_y^2 + S_z^2 \end{equation} satisfies: \begin{align} S^2 |1,1\rangle & = 2 \hbar^2 |1,1\rangle \\ S^2 |1,0\rangle & = 2 \hbar^2 |1,0\rangle \\ S^2 |1,-1\rangle & = 2 \hbar^2 |1,-1\rangle \\ S^2 |0,0\rangle & =0|0,0\rangle \end{align} Therefore, we see that the following two important equations are satisfied: \begin{equation} \begin{array}{cc} S^2 |s , m \rangle = \hbar^2 s (s+1) |s , m \rangle \; ,& S_z |s , m \rangle = \hbar m |s , m \rangle \end{array} \end{equation} To sum up, we have found the possible eigenvalues for the magnitude and $z$-component of the system and the eigenstates corresponding to these values: the allowed values for total spin are $s=1$ and $s=0$, while the allowed value of $s_z$ are $\hbar$, $0$, and $-\hbar$ and the corresponding eigenstates in the product basis $|1,1\rangle$, $|1,0\rangle$, $|1,-1\rangle$ and $|0,0\rangle$. Thus, the meaning of these triplet and singlet states is that they are the possible states of the system consisting of the two aforementioned particles. This is often written as: \begin{equation} \frac{1}{2} \otimes \frac{1}{2} = 1 \oplus 0 \end{equation} which means that the tensor product of two spin-$1/2$ Hilbert spaces is a direct sum of a spin-$1$ space and a spin-$0$ space.