Why are only some eigenstates observed upon collapse?

If you have the first state and measure the vertical spin, it will always give +1, but if you measure it horizontally, it will result in +1 or -1 with equal chance, and collapse to one of those states.

By the postulates of QM, the state vector after measuring an observable is the projection of the state vector before the measurement onto the eigenspace corresponding to the result of the measurement, which is guaranteed to be an eigenvalue of the operator associated with that observable. Furthermore, the probability of collapsing into any given eigensystem is determined by "how much" the initial state projects onto that eigensystem.

So mathematically, the answer is because that's what we defined the theory to do. The state vector collapses to an eigenstate of the observable you measured, and that measurement has a probability associated with it that is determined by the initial state. Physically, I would say that's up to the interpretation of QM you want to go with, but physics can't really answer why this is ultimately.

It is true that you can express the "eigenstates" in terms of other "eigenstates", but here these correspond to different operators. For both given states, measuring vertically will give back the same state because these correspond to the (say) $sigma_{z}$ [vertical] eigenstates. But, measuring horizontally will give an equal probability for measuring left or right since the up state is NOT an eigenstate of (say) $sigma_{x}$ [horizontal]. The wavefunction will "collapse" to one of the eigenstates of the measured operator.