Is this definition of orthohelium and parahelium incorrect? 
"One electron is presumed to be in the ground state, the 1s state. An electron in an upper state can have spin antiparallel to the ground state electron ($S=0$, singlet state, parahelium) or parallel to the ground state electron ($S=1$, triplet state, orthohelium)."
From HyperPhysics

When they say "parallel to the ground state electron" then is it assuming that they are both spin up, or both spin down? If so, isn't it then ignoring the $S=1$ state with spin up and spin down:
$$|1\rangle |0\rangle = \frac{|+\rangle|-\rangle + |-\rangle|+\rangle}{\sqrt 2}$$
Therefore, if one electron is presumed to be in the ground state, 1s, state, if the spins can be opposite, a second electron can also occupy the ground state in $S=1$ orthohelium. Is this correct?
 A: The spins are not antiparallel in this state as @my2cts answer correctly states (as demonstrated by $\vec S_1 \cdot \vec S_2$ having a positive eigenvalue).
I was confused by this at first too, especially the argument, that the measurements of $S_{1z}$ and $S_{2z}$ will point to opposite directions due to entanglement seemed convincing to me.
But it just looks that way because the $S_z$ basis is the wrong basis to look at. They have to point in opposite directions in this basis, because the total $m_z = 0$. So you are just asking a spin state in the $x$-$y$-plane for its $z$-component. This way you can't find out whether the spins are parallel or antiparallel in that plane.
To make the parallel orientation of the spins manifest in the expression for the state, you'll have to rewrite the state in a basis in the plane.
We rewrite our state terms of the eigenstates $\lvert \pm \rangle$ of $S_x$: $$ S_x \lvert \pm \rangle = \pm \frac \hbar 2 \lvert \pm \rangle$$
In terms of those states we have
$$ \lvert \uparrow / \downarrow \rangle = \frac{1}{\sqrt{2}} \big( \lvert + \rangle \pm \lvert - \rangle \big)$$
in terms of these states the two product states become
$$ \lvert \uparrow \downarrow \rangle = \frac 1 2  \big( \lvert + \rangle + \lvert - \rangle \big) \otimes \big( \lvert + \rangle - \lvert - \rangle \big) = \frac 1 2 \big( \lvert++\rangle - \lvert+-\rangle + \lvert -+ \rangle - \lvert -- \rangle \big) $$
$$ \lvert \downarrow \uparrow \rangle = \frac 1 2 \big( \lvert++\rangle - \lvert-+\rangle + \lvert +- \rangle - \lvert -- \rangle \big) $$
So the triplet state with $m_z = 0$ is given by the following in this basis:
$$ \frac{\lvert \uparrow \downarrow \rangle + \lvert \downarrow \uparrow \rangle}{\sqrt{2}} = \frac{\lvert ++ \rangle - \lvert -- \rangle}{\sqrt{2}}$$
And in this form we can see manifestly that the spins are in parallel. (Similarly, this could be done for the eigenstates of $S_y$.)
A: With such a two-electron spin wave function, both electrons cannot be in the 1s state due to the electron parity. (The wave function should be antisymmetric relative to exchange of two particles.) When both electrons are in the 1s orbital, the only possible spin function reads
$$\frac{1}{\sqrt{2}} \left( |\uparrow \downarrow \rangle-| \downarrow \uparrow\rangle \right)$$
A: The definition of orthohelium as having two parallel electron spins is correct. In the state $$\frac1{\sqrt2}\big(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle\big)$$ the two spins are also parallel. It may seem that they are not, but note that this state has $m_z=0$, so both spins are perpendicular to the quantization direction. Indeed, the operator $S_+$ will turn this state into $$\big(|\uparrow\uparrow\rangle\big)$$.
Also, this is an eigenstate of $S^2$ with eigenvalue 2$\hbar^2$, just like the other members of the triplet, while $S_1 \cdot S_2$ has eigenvalue $+\frac{\hbar^2}{4}$. The positive sign indicates that the spins are parallel. Indeed for the singlet this eigenvalue is $-\frac{3\hbar^2}{4}$, which indicates antiparallel alignment.
A: Yes you can do that; the space and spin parts just have to have opposite symmetry characteristics so that the total wavefunction is antisymmetric.
