Term symbols: counting electrons. How?
This example is probably described quite thoroughly on the internet for the case of $\ce{O2}$ (which is isoelectronic), so it may be worth searching a bit for alternate explanations.

You cannot have a $^3\Delta$ term symbol, because of the Pauli exclusion principle. The π* orbitals come as a set of two: essentially, one of those two is associated with +1 angular momentum, and the other is associated with −1 angular momentum.

Here, to be specific, "angular momentum" refers to the projection of the angular momentum onto the internuclear axis. The "angular momentum" quantum number of each MO is denoted $\lambda$, so in other words, the π* orbitals come as a set of $\lambda = +1$ and $\lambda = -1$.

The quantum number that measures the "angular momentum" of the term is denoted $\Lambda$, and because $\Lambda$ is a projection quantum number, the value of $\Lambda$ is simply the sum of the individual $\lambda$ values. The Greek letter that represents it in the term symbol depends on the absolute value of $\Lambda$: if $\Lambda = 0$ then it is a $\Sigma$ term, if $|\Lambda| = 1$ then it is a $\Pi$ term, and if $|\Lambda| = 2$ it is a $\Delta$ term.

Now, in the ground state configuration of $\ce{S2}$, we can ignore all the orbitals below the π* orbitals because all those $\lambda$'s sum to zero. The only thing of interest is the π* electrons.

If you want to have a triplet state, i.e. the electrons have the same spin, then you cannot put them into the same π* orbital: one will have to go into the $\lambda = +1$ orbital, and the other will have to go into the $\lambda = -1$ orbital. Hence you get $\Lambda = 0$, which leads to a $^3\Sigma$ term symbol, not a $^3\Delta$.

Finding the parity of the term symbol is nothing more than multiplying all the individual parities of the electrons (i.e. for doubly occupied $\mathrm{g}$ MOs you need to multiply by $\mathrm{g}$ twice). This is useful, because $\mathrm{g \otimes g = u \otimes u = g}$, which allows you to immediately discard all doubly occupied MOs. You only need to look at the two π* electrons, which are both $\mathrm{u}$, leading to an overall $\mathrm{g}$ parity and a $^3\Sigma_\mathrm{g}$ term symbol.

For $\Sigma$ term symbols there is an additional detail in that you need to specify the symmetry with respect to reflection in a plane containing the molecule. In this case it turns out that the molecular wavefunction is antisymmetric with respect to this reflection, so the proper term symbol is $^3\Sigma_\mathrm{g}^-$.