Define $f:S^2\rightarrow\mathbb{R}$ by $f(x,y,z) = z$. What are the orientations on the preimage circles $S(t) = f^{-1}(t)$ for $t\in(-1,1)$? (Explicitly exhibit a positively oriented tangent vector at an arbitrary point).
The first thing that was never really made clear to me: for any point $r\in \mathbb{R}$, is $T_{r}\mathbb{R}$ the positive zero space? or does the sign correspond to the sign of $r$ somehow, if so why?
Then, how would one go about getting an orientation for $S^2$? Is there a standard one?
Any hints whatsoever would be greatly appreciated.
First things first. You should be aware that $f$ isn't a submersion everywhere, because you have 2 critical points (at $z=-1$ and $z=1$). So when you take $S(t)$, you must have $-1<t<1$ in order to get a nice differentiable structure.
Now about orientations.
You should now that there is a standard orientation of $\mathbb R^1$ given by the canonical vector $(1)$. (Intuitively, It's pointing to the right.)
For any point $r\in \mathbb R$, the tangent space at $r$ is strictly the same as in $r=0$. (Because you can translate $r$ to $0$, without changing anything when taking derivatives.) I'm not sure about what you mean when you say "positive zero space", but you should be aware that $T_r\mathbb R$ is isomorphic to $\mathbb R$. You certainly can give an orientation for each $T_r\mathbb R$, but you have the choice of the positive or negative one. This choice is the same as taking an orientation on $\mathbb R^2$.
There is also a standard orientation for $S^2$. There is many way to orient $S^2$ given the standard orientation of $\mathbb R^n$. For example the diffeomorphism $S^2-\{a\}\to \mathbb R^2$ given by stereographic projection, is a way to define it.