I wrote the following argument to prove that $S^1$ is parallelizable, that is, to show that the tangent bundle is trivial. It looks fairly reasonable to me.
Let $\tau=2\pi$.
We define a map $\varphi:S^1\times\Bbb R\to TS^1$ by $\varphi((e^{i\tau\theta}, t))=(e^{i\tau\theta},t\frac{\partial}{\partial x^1}\Big|_{e^{i\tau\theta}})$. This map is clearly a bijection: injectivity follows trivially and surjectivity follows since the tangent space $T_{e^{i\tau\theta}}$ is one-dimensional.
It remains to be shown that it is a diffeomorphism. We will, in fact, show that it is the identity map on suitably chosen coordinates. For any point $(e^{i\tau\theta},t)\in S^1\times\Bbb R$, we can choose the coordinate chart such that $(e^{i\tau(\theta+\varepsilon)},(t+\delta))$, for sufficiently small $\varepsilon$ and $\delta$, is given by $(\varepsilon,\delta)$.
Similarly, for any point $(e^{i\tau\theta},t\frac{\partial}{\partial x^1}\Big|_{e^{i\tau\theta}})\in TS^1$, we can choose the coordinate chart such that $(e^{i\tau(\theta+\varepsilon)},(t+\delta)\frac{\partial}{\partial x^1}\Big|_{e^{i\tau(\theta+\varepsilon)}})$, for sufficiently small $\varepsilon$ and $\delta$, is given by $(\varepsilon,\delta)$. Then, we have that $\widehat\varphi$ (the function with respect to these two coordinates) sends $(\varepsilon,\delta)\mapsto (\varepsilon,\delta)$, as desired.
However, I don't see where this is using the tangent bundle construction in any meaningful way. It seems that this would work just as well for any rank-1 vector bundle over the sphere.
That concerns me, because of course the conclusion is not true: for instance the Möbius strip is (diffeomorphic to) a vector bundle over the sphere. Since it is not orientable, we know that the bundle is nontrivial.
So the question is: Where does this break for a general vector bundle? Or if the argument is simply invalid, can it be fixed without much trouble?
You're implicitly using the fact that a non-vanishing section (the coordinate vector field $\partial/\partial x_{1}$) exists. The tangent bundle admits such a section, while the non-trivial line bundle doesn't.