I was recently surprised to learn that $TS^2$ is not trivial. I had a few questions regarding this:
For all $x\in S^2$ does there exist an open set $U$ containing $x$ such that $TU$ is trivial? If $(U',\phi$) is a chart of $S^2$ with $x\in U'$ is it true that that $TU'$ is trivial? What if we replace $S^2$ with any manifold?
I also had a more open ended question. How do you visualise $TS^2$? What geometric property of $TS^2$ makes it so that we can't just say it's the same as $S^2\times \mathbb R^2$?
There always exists an open cover $\{U_\alpha\}$ of $M$ such that $TU_\alpha$ is homeomorphic to $U_\alpha \times \mathbb{R}^{\text{dim } M}$. The collection of these maps along with the open cover is called a local trivialization, and it can be done for the tangent bundle of any manifold.
I can't say I have a way to visualize the space $TS^2$, but to answer your question about what property makes it non trivial consider this: On $S^2\times\mathbb{R}^2$ there is a non vanishing "section" or vector field, one can pick $X(p)=(p,e_1)$ for any $p\in S^2$ and $e_1$ is the first standard basis vector of $\mathbb{R}^2$. However one cannot find a non vanishing vector field on $S^2$ (this is known as the hairy ball theorem) which tells us that $TS^2$ cannot be trivial.