Why can't $\frac{d}{d \theta}$ be considered as a globally defined vector field on $S^1 \subset \mathbb{R}^2$?
Lee's Introduction to Smooth Manifoolds, page 176 Second edition, Example 8.4:
Let $\theta$ be any angle coordinate on a proper open subset $U \subset S^1$ and let $\frac{d}{d\theta}$ denote the corresponding vector field. Because any other angle coordinate $\gamma$ differs from $\theta$ by an additive constant in a neighborhood of each point, the transformation law for coordinate vector fields shows that $\frac{d}{d\theta}=\frac{d}{d\gamma}$ on their common domain. For this reason, there is a globally defined vector field on $S^1$ whose coordinate representation is $\frac{d}{d\theta}$ with respect to any angle coordinates. It is a smooth vector field because its component function is constant in any such chart. We denote this global vector field by $\frac{d}{d\theta}$, even though, strictly speaking, it cannot be considered as a coordinate vector field on the entire circle at once
Can one of you genius's comment on the part I put in bold? Why can't $\frac{d}{d\theta}$ be considered as a coordinate vector field on the entire circle at once?
This is because $\frac{d}{d\theta}$ is only defined on $U$, the coordinate chart. Since there is no global coordinate chart for $S^1$, you can’t build a vector field at every point pulling back a single tangent vector from $\mathbb{R}$.