My feeling is that you are confused by notation.
I read that when I have two identical particles with spin 1/2 there are 4 possibilities
These are not four possibilities, but these are four basis states. They form a complete orthonormal set of states for two spins. These basis states span the Hilbert space of all possible two-spin states.
The notation, for example $\left|\left.\uparrow\downarrow\right>\right.$, is a shorthand for $$ \left|\left.\uparrow\downarrow\right>\right. \equiv \left|\left.\uparrow\right>\right._1\otimes\left|\left.\downarrow\right>\right._2,$$ meaning that particle 1 is in state $\left|\left.\uparrow \right>\right.$ and particle 2 is in state $\left|\left.\downarrow\right>\right.$ (note, there is a logical and).
what's the difference between these 2 states $$ \left|\left.\uparrow\downarrow\right>\right., \left|\left.\downarrow\uparrow\right>\right.$$
Given the above, you can now tell the difference: in the first case, particle 1 is in state $\left|\left.\uparrow\right>\right.$ and particle 2 is in state $\left|\left.\downarrow\right>\right.$, while in the second case particle 1 is in state $\left|\left.\downarrow\right>\right.$ and particle 2 is in state $\left|\left.\uparrow\right>\right.$.
So far, we made no use of the indistinguishability of the two particles, we even labeled them as particle 1 and particle 2, so we treated them as distinguishable particles.
Then since there is the symmetrization requirement I can take as eigenvalues the following 4 states: $$ \left|\left.\uparrow\uparrow\right>\right., \left|\left.\downarrow\downarrow\right>\right.,\frac{1}{\sqrt{2}}\left(\left|\left.\uparrow\downarrow\right>\right.+\left|\left.\downarrow\uparrow\right>\right.\right), \frac{1}{\sqrt{2}}\left(\left|\left.\uparrow\downarrow\right>\right.-\left|\left.\downarrow\uparrow\right>\right.\right).$$
Note first that I have put factors of $1/\sqrt{2}$ instead of your factors $1/2$. This is, because we want the states to remain normalized. Second, please be aware that these four states are in no way eigenvalues in this context, they are also not eigenstates, because you did not specify any hamiltonian, of which these states could be the eigenstates. 66
These four new states simply form an alternative complete and orthonormal set of basis states for two spins.
Now let me try to explain, why these four states are well suited to describe indistinguishable particles. To have indistinguishable particles should mean that it should not matter in our description of the two particles, which of them is in which state. In other words, when we exchange the labels 'particle 1' and 'particle 2', we should get the same observable properties of the system. This is indeed the case for our four new basis states. If you remember our notation convention from above, then, for example, $$ \left|\left.\psi\right>\right. = \frac{1}{\sqrt{2}}\left(\left|\left.\uparrow\downarrow\right>\right.-\left|\left.\downarrow\uparrow\right>\right.\right) = \frac{1}{\sqrt{2}}\left(\left|\left.\uparrow\right>\right._1\otimes\left|\left.\downarrow\right>\right._2-\left|\left.\downarrow\right>\right._1\otimes\left|\left.\uparrow\right>\right._2 \right).$$ Now exchanging the labels for particle 1 and 2, we have $$ \frac{1}{\sqrt{2}}\left(\left|\left.\uparrow\right>\right._2\otimes\left|\left.\downarrow\right>\right._1-\left|\left.\downarrow\right>\right. 66