Fundemental group of $S^2$ with equator is identified $z \sim z^3$

Take the northern hemisphere $N$ and the southern hemisphere $S$. Identify $z \in \partial N$ with $z^3 \in \partial S$. Then the path $c(t) = \exp(\frac{2\pi i}{3} t), t \in [0, 1]$, in the northern hemisphere gets identified with the path $h(t) = \exp(2\pi i~t)$ in the southern hemisphere which is a loop. In fact, it's a loop that traverses the equator of the southern hemisphere. Imagine that it's a rubber band, and it's attached to $S$ at the point $z = 1$, but everywhere else it's coated in grease. Now take the part of the rubber band near $z = -1$ and press it southward. It'll slip down, pass over the south pole, and eventually contract up to the point that you've held fixed at $z = 1$. And that's your homotopy.

I think you are getting confused because the space you describe (which I'll call $X$) really has two different "equators". Note that $X$ is not simply the quotient of $S^2$ by an equivalence relation on the equator. Instead, $X$ is two hemispheres attached together where $z$ in the equator of one hemisphere is identified with $z^3$ in the equator of the other hemisphere.

So, this does mean that a path on the first equator which goes around an arc of $2\pi/3$ is a loop, since its endpoints become identified when you glue it to the second hemisphere. There is no nullhomotopy of this loop inside the first hemisphere. But when you look at this loop as living in the second hemisphere instead, it's just an ordinary loop that goes around the equator once. So, it is nullhomotopic inside the second hemisphere, which really is just homeomorphic to an ordinary disk (unlike the first hemisphere whose equator has gotten glued together).

(If you instead had a space which is the quotient of $S^2$ by the equivalence relation $z\sim e^{2\pi i/3}z$ on the equator, then it would not be simply connected and indeed the loop going a third of the way around the equator would not be nullhomotopic.)