I am finding it difficult to understand the four states of two entangled qubits, $$\left|00\right>$$ $$\left|01\right> + \left|10\right>$$ $$\left|01\right> - \left|10\right>$$ $$\left|11\right>$$
This Veritasium video states that two qubits have these four states, and calls the $\left|01\right> + \left|10\right>$ the $\left|T_0\right>$ state, and $\left|01\right> - \left|10\right>$ the singlet state, $\left|S\right>$.
However, what does the difference between the minus and plus signify? I understand that in both states the qubits have the property of being opposite to one another, but that's it.
It would be greatly appreciated if someone could help me understand this difference!
Before I mark an answer as correct, the main understanding I now have, is that, although when squared having an identical probability, the states differ when transformations are applied to it?
Is this basically like this image, taken from here?
Note; I think my question differs from this post. The OP of that post already understands the presence and meaning of the minus sign, but rather asks why the minus sign is associated with $S=0$, which is a different question. Yet this question has been marked as duplicate, without any argument why it is still the same. I would like to see an argument against.