Coupling of two spin half particles If there are two electrons coupled by interaction having hamiltonian H=A*S1*S2 where S1 and S2 are spin angular momentum operators of two electrons, we know we have four possible eigenstates for the combined system. The diagram here shows the possible situations. I have following questions:
1)- Two of the four states are combinations of pure states, how can they be eigenstate because in eigenstate we should be able to get exact values upon measurement which is not the case here?
2)- What causes the independent electrons to couple? (Non-mathematical explanation)
3)- Since S1 , S2 are spin angular momentum vector operators , they can have any direction on the cone about z-axis independent of other electron, but in the diagram shown we can see that only 4 combinations of **S1 and S2 ** are allowed. What is reason for only these orientations of spin momentum ? For example can't we have a situation where both are on upper part of the cone with tips diametrically opposite ? Can someone derive these four cases which are allowed ?
Can someone explain these things without too much maths and by giving physical reasons ?
 A: Let’s address your third part first as that’s the fundamental one. 

What is reason for only these orientations of spin momentum ?

It all stems from the fact that electrons are indistinguishable. This means that if you interchange the (any) two electrons, the observables must not change. This means in your wavefunction, if you swap the two individual spins the orthonormality of the wavefunctions must remain be preserved. Let us look at two cases in specific:
$$|{\psi_{12}}\rangle=|{\uparrow\uparrow}\rangle\to |{\uparrow\uparrow}\rangle = |{\psi_{12}} \rangle\\
|{\psi_{12}} \rangle =\frac{1}{\sqrt{2}}\Big(|{\uparrow\downarrow}\rangle - |{\downarrow\uparrow}\rangle\Big)\to \frac{1}{\sqrt{2}}\Big(|{\downarrow\uparrow}\rangle - |{\uparrow\downarrow}\rangle\Big)= -|{\psi_{12}} \rangle $$
These states at most get an overall minus sign that doesn’t matter for inner products. Whereas a state like:
$$|{\psi_{12}}\rangle=|{\downarrow\uparrow}\rangle\to |{\uparrow\downarrow}\rangle \ne \pm|{\psi_{12}} \rangle$$
Doesn’t preserve norms under a particle swap. So these states are the only possible states due to the indistinguishable nature of electrons. 
Note that the two cones represent two independent spins, thus two arrows can’t be on the same cone in that depiction. 

Two of the four states are combinations of pure states, how can they be eigenstate because in eigenstate we should be able to get exact values upon measurement which is not the case here?

Your Hamiltonian is of the form $S_1\cdot S_2$. So your energy eigenstates must be the eigenstates of this operator and not $S_1$ or $S_2$ individually. And you can easily check that the 4 states listed are eigenstates of the given Hamiltonian. These are the eigenstates when you measure energy. 

What causes the independent electrons to couple? (Non-mathematical explanation)

Independent electrons in the presence of each other are no longer independent and interact with each other. This causes them to couple. Interestingly these couplings under the right conditions allow the system of electrons to couple and behave as a boson rather than a fermion that allows for properties like superconductivity and such. 
A: The interaction can have different origins. The expression you give $$H=A \vec S_1 \cdot \vec S_2$$ is an effective spin Hamiltonian. It us used when only the spin degrees of freedom are relevant . The interaction can be an exchange interaction or a hyperfine interaction. Exchange interaction  results from Coulomb interaction of a pair of identical fermions, due to the antisymmetry under particle exchange of the wave function. Hyperfine interaction results from interaction of the two electron magnetic moments.
