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.
