For a two-level state space, with a basis $\{|0\rangle,|1\rangle\}$, the infinitessimal rotation generators are
$$
J_j = \frac{1}{2}\sigma_j
$$
where $\sigma_j$ for $j = 1,2,3$ are the three Pauli matrices.
If you want to rotate a state around the $x$-axis by the angle $\theta$, you can multiply the state by the $2 \times 2$ matrix $e^{-i \theta J_1}$.
When you tensor together two of these two-level state spaces, you get a four-dimensional state space with a basis $\{|00\rangle,|10\rangle\,|01\rangle,|11\rangle\}$, the infinitessimal rotation generators become
$$
J_j = \frac{1}{2}\sigma_j\otimes I+\frac{1}{2}I \otimes \sigma_j
$$
where $I$ is the identity matrix.
You can explicitly check that for the state
$$
|\psi\rangle = \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle) = \frac{1}{\sqrt{2}}(|0\rangle \otimes |1\rangle - |1\rangle \otimes |0\rangle)
$$
that
$$
J_j |\psi\rangle = 0
$$
for all $j = 1,2,3$. Therefore when you perform a rotation on this particular state $|\psi\rangle$, the state does not change. This means $|\psi\rangle$ transforms in the trivial representation of the rotation group, which we call "spin $0$."