All the three states correspond to the angular momentum of magnitude 1, i.e. $S^2 = \hbar^2 n(n+1)$, where $n=1$. The projection of this angular momentum on $z$-axis can take three values: $S_z = 0,\pm \hbar$.
It is easy to see which state corresponds to which projection by acting on them with the operator of total spin projection on $z$-axis, $\hat{S}_z = \frac{\hbar}{2}(\hat{\sigma}_z^{(1)} + \hat{\sigma}_z^{(2)})$:
\begin{array}
\hat{S}_z |\uparrow\uparrow\rangle = (\frac{\hbar}{2} + \frac{\hbar}{2})|\uparrow\uparrow\rangle = \hbar|\uparrow\uparrow\rangle,\\
\hat{S}_z |\downarrow\downarrow\rangle = (\frac{\hbar}{2} + \frac{\hbar}{2})|\downarrow\downarrow\rangle = \hbar|\downarrow\downarrow\rangle,\\
\hat{S}_z \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle) =
(\frac{\hbar}{2} - \frac{\hbar}{2} - \frac{\hbar}{2} + \frac{\hbar}{2}) \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle) = 0.
\end{array}
The state with zero projection on the $z$-axis is very similar to the singlet state, which has zero momentum and zero projection on the $z$-axis, and the wave function $$\frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle).$$
In my opinion, the easiest way to get the grasp of this is by diagonalizing the Hamiltonian for two interacting spins:
$$\hat{H} = \hat{\vec{S}}_1\cdot \hat{\vec{S}}_2 =
\hat{S}_{1x}\hat{S}_{2x} + \hat{S}_{1y}\hat{S}_{2y} + \hat{S}_{1z}\hat{S}_{2z}$$
in the basis $|\downarrow\downarrow\rangle, |\uparrow\downarrow\rangle, |\downarrow\uparrow\rangle,|\uparrow\uparrow\rangle$.