Singlet and Triplet states: Why is the $S=0$ state defined as it is?

You are correct that one of the $S = 1$ states has zero angular momentum along the $z$ axis. However, the $S = 0$ state has zero angular momentum along any direction, and this follows directly from the presence of the minus sign.

To see why, consider the operator for spin along a general direction, $S_{\hat{n}}$. Since the spins are identical, it doesn't matter which spin is what, so $$\langle \uparrow \downarrow | S_{\hat{n}} | \uparrow \downarrow \rangle = \langle \downarrow \uparrow | S_{\hat{n}} | \downarrow \uparrow \rangle = \alpha(\hat{n}).$$ Moreover, the operator $P$ that interchanges the spins doesn't affect $S_{\hat{n}}$, because $$P S_{\hat{n}} = P \left(S_{\hat{n}}^1 + S_{\hat{n}}^2 \right) = S_{\hat{n}}^2 + S_{\hat{n}}^1 = S_{\hat{n}}$$ which further implies that $$\langle \uparrow \downarrow | S_{\hat{n}} | \downarrow \uparrow \rangle = \langle \downarrow \uparrow | S_{\hat{n}} | \uparrow \downarrow \rangle = \alpha(\hat{n}).$$ This is all the information we need to evaluate the spin along the $\hat{n}$ direction for the two states you give. For the $S = 1$ state, we find $$\frac{\alpha}{2} + \frac{\alpha}{2} + \frac{\alpha}{2} + \frac{\alpha}{2} = 2\alpha$$ while for the $S = 0$ state we find $$\frac{\alpha}{2} + \frac{\alpha}{2} - \frac{\alpha}{2} - \frac{\alpha}{2} = 0.$$ The cross terms pick up a minus sign, due to the minus sign in the superposition. This shows the singlet has zero spin along any direction.

It's important to remember that when we say $S=1, 0$ we're really thinking of the eigenstates and eigenvalues of $\mathbf{S}^2 = (\mathbf{S}_1 + \mathbf{S}_2)^2$. Try applying this operator, and you'll see that the symmetric states have eigenvalue $S=1$, and the antisymmetric will have eigenvalue $S=0$. If you're simply applying $\mathbf{S}_z$, which is what you are doing in your question statement, then definitely both the mixed symmetric and antisymmetric states have eigenvalue $j=0$, as you've pointed out.