Checking that $f_{\alpha}$ is a bundle isomorphism is the same as checking two things
$f_{\alpha}$ is a diffeomorphism ;
$f_{\alpha}$ commutes with the projection maps $\pi_{\alpha} : \mathrm{U}_{\alpha} \times \mathbb R^n \to \mathrm{U}_{\alpha}$ and $\pi'_{\alpha} : \mathrm{TM}_{|\mathrm U_{\alpha}} \to \mathrm{U}_{\alpha}$ i.e you have to check that $f_{\alpha} \circ \pi_{\alpha} = \pi'_{\alpha} \circ f_{\alpha}$.
By definition of the tangent bundle $\mathrm{TM}_{|\mathrm U_\alpha} = \bigcup_{x \in \mathrm U_{\alpha}} \{x\} \times \mathbb R^n$
We deduce that $f_{\alpha}$ is a diffeomorphism. The only thing left to check is the compatibility with the projection maps. But this comes immediately.
Hence $f_{\alpha}$ is a bundle isomorphism.