Why does the Gibbs Free Energy change determine spontaneity of a reaction?
By the second law of thermodynamics, your main goal is to ensure that $$\Delta S_{\mathrm{universe}} \geq 0$$

$$\Delta S_{\mathrm{universe}} = \Delta S_{\mathrm{sys}} + \Delta S_{\mathrm{external}}$$

$\Delta S_{\mathrm{sys}}$ is measurable on your system. $\Delta S_{\mathrm{external}}$ is a bit more challenging, since you're not going to count the microstates of the external system, so you compute indirectly by heat transfer to the surrounding. Thus,

$$\Delta S_{\mathrm{external}} = -\frac{\Delta H}{T}$$

Combine:

$$T\Delta S_{\mathrm{universe}} = T\Delta S_{\mathrm{sys}} - \Delta H$$

or

$$\Delta G = -T\Delta S_{\mathrm{universe}} = \Delta H -T\Delta S$$

And with the sign flip, you're looking for $\Delta G \leq 0$.

Importantly, I should point out that this is not a closed system in the sense that we assume constant-pressure and not constant-volume.

If you're looking at a process that is constant-volume, you need to replace $\Delta H$ (which has pressure-volume work compensation) with $\Delta U$ (which does not) and replace $\Delta G$ (Gibbs free energy, constant pressure) with $\Delta F$ (Helmholtz free energy, constant volume).