Much of this discussion was based on the text “Geometric Control of Mechanical Systems” by Francesco Bullo and Andrew D.Lewis. A more formal development of the following ideas can be found in that text, this is simply meant to be a page of informal notes.
Table of Contents
Equivalence Classes
Consider the equivalence class $\mathcal{R}\equiv\mathbb{Z}$. Then $\mathbb{R}/\mathbb{Z}$ can be visualized as 
Consider $\mathbb{S}^1$ as the unit circle in the complex plane. Next, consider the surjective function $\varphi:\mathbb{R}\to\mathbb{S}^1$ as $\varphi(x) = e^{2\pi ix}$. Note that $\varphi$ is a Group Homomorphism, and that $Ker(\varphi)=\mathbb{Z}$. Therefore, under the first group isomorphism theorem, we have that $\mathbb{R}/\mathbb{Z}\cong\mathbb{S}^1$. Imprecisely, this can be visualized in the following way
Tangent Bundles
Consider the curve $\gamma:I\to M$, with $\gamma’(x)$ representing the derivative of $\gamma$ in an appropriate chart at the point $x$. The tangent fiber at a point x is the set of all curves $\gamma$ that have the same derivative at $x$, denoted by $T_x M$. The tangent bundle $TM = \sqcup_{x\in M}T_x M$, or the disjoint union of the tangent fibers as seen in
A section of the tangent bundle $\xi\in\Gamma(M)$ provides a vector field on $M$.
Nontrivial Vector Bundle
Let $M=[0,1]\subset\mathbb{R}$ be an embedded submanifold. For the proceeding discussion, let $\pi:V\to M$ be the projection mapping from the vector bundle $V$ to the base manifold. Now consider the equivalence relation $\mathcal{R}_1=\{((0,y),(1,y)):y\in\mathbb{R}\}$, as visualized by
The vector bundle $V_1=M\times \mathbb{R}/~$ is isomorphic to $TM$ because of the parallelizable nature of $M$, and is a cylinder as seen in
If instead we consider the equivalence class $\mathcal{R}_2=\{((0,y),(1,-y)):y\in\mathbb{R}\}$
the vector bundle $V_2=M\times\mathbb{R}/\sim$ is the infinite Möbius band, as seen in
Note that $\pi(V_i)\cong\mathbb{S}^1$ for $i=\{1,2\}$.