For an elliptic curve $E$, I saw a paper once mention that you can interpret a rational point on $E$ as a degree $0$ Galois cohomology class. I'm familiar with group cohomology but not all that much with Galois cohomology. I know that you can associate absolute Galois groups with torsion points of an elliptic curve, but that doesn't seem to immediately give the above result.
How is that statement true?
Galois cohomology is simply group cohomology for the absolute Galois group that accounts for the profinite topology (i.e. you require cocycles to be continuous with respect to the topology). With that said, for any module with a continuous action of $\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$, the $0^\text{th}$ cohomology group is simply the fixed points of this action. In particular, $$H^0(\mathbf{Q}, E) := H^0(\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}), E(\overline{\mathbf{Q}})) = E(\overline{\mathbf{Q}})^{\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})} = E(\mathbf{Q}).$$
I'm not sure what you are referring to when you say that you can associate absolute Galois groups with torsion points. The torsion points simply have an action of the absolute Galois group.