Up and down spin are orthogonal, not antiparallel In conventional coordinate systems (anything you solve a simple Newtonian mechanics problem with), up and down are + and - z. A vector pointing up and a vector pointing down are anti-parallel.
But in qm, we have up and down spinors making and orthonormal basis. These basis vectors are also called positive and negative z spin angular momentum. I understand the math for how spinors like (1,0) and (0,1) are orthogonal. I also see how they can be expressed as superpositions of x and y spinors, using complex numbers such that a two-component spinor can represent quantities in 3 dimensions.  (this seems a little like what I have studied about symmetry groups like SU(1), so if that is relevant in the solution, I appreciate a discussion, but if it is unrelated, please do not bother correcting any huge mistakes in this sentence because I did not fully try to study it on my own yet).
My question is this: what is the intuition for saying that spin up and down are orthogonal?
 A: The inner product of two spin vectors is not spatial overlap. Rather, $|\langle \psi_1 | \psi_2 \rangle |^2$ is the probability of measuring the state in $| \psi_2 \rangle$ if it is originally in $|\psi_1 \rangle$. Therefore, if a state is spin up, $(1,0)$, then you have  $0\%$ chance of measuring it to be spin down, $(0,1)$.
A: An addendum to Thomas Fritsch's answer that resolves much of the initial confusion: you measure the eigenvalue of the operator acting on the state, not the eigenvector for the state. The relevant spin operator on the eigenvector clearly gives the desired anti-parallel eigenvalues, and the eigenvectors are simply vectors which behave in the desired manner, without an equally direct geometric interpretation (discussed here).
Some intuition for the orthogonality is: the z spin observed in a measurement must be +- 1/2, it cannot be zero. The orthogonality indicates that a body spinning upwards does not need to be expressed in a nontrivial sum of "up-ness" and down-ness." It is expressed solely as down-ness and has zero up-ness. This indicates orthogonality.
A: You need to distinguish the orthogonality in spinor space ($\mathbb{C}^2$)
from orthogonality in vector space ($\mathbb{R}^3$).
The spaces are different, and therefore scalar product and orthogonality
in these spaces have entirely different meanings.
Example:
The two spinors
$$|\uparrow\rangle=\begin{pmatrix}1\\0\end{pmatrix}$$
and
$$|\downarrow\rangle=\begin{pmatrix}0\\1\end{pmatrix}$$
are orthogonal to each other because their scalar product is zero:
$$\langle\uparrow|\downarrow\rangle=0$$
The expectation values of the spin vector $\vec{S}=\frac{\hbar}{2}\vec{\sigma}$
(where $\vec{\sigma}$ is the Pauli vector
$\vec{\sigma}=\sigma_x\hat{x}+\sigma_y\hat{y}+\sigma_z\hat{z}$)
for these two spinors are:
$$\vec{S}_\uparrow
= \langle\uparrow|\vec{S}|\uparrow\rangle
= \frac{\hbar}{2}\langle\uparrow|\vec{\sigma}|\uparrow\rangle
= + \frac{\hbar}{2} \hat{z}$$
and
$$\vec{S}_\downarrow
= \langle\downarrow|\vec{S}|\downarrow\rangle
= \frac{\hbar}{2}\langle\downarrow|\vec{\sigma}|\downarrow\rangle
= - \frac{\hbar}{2} \hat{z}$$
These two vectors are antiparallel to each other.
Their scalar product $\vec{S}_\uparrow \cdot \vec{S}_\downarrow$ is not zero.
