The pullback of the Euclidean tangent bundle is canonically trivial

The identity $\mathrm{id}:{\mathbb{R}}^r\to {\mathbb{R}}^r$ is a smooth chart whose domain is all of ${\mathbb{R}}^r$, so by [F1] its induced tangent-bundle chart $\stackrel{}{\mathrm{id}}:T{\mathbb{R}}^r\to {\mathbb{R}}^r\times {\mathbb{R}}^r$, $v\mapsto (p,v^1,\dots ,v^r)$, is a diffeomorphism onto ${\mathbb{R}}^r\times {\mathbb{R}}^r$; it is linear on every fibre. Hence it is a smooth bundle isomorphism $T{\mathbb{R}}^r⟶{\epsilon }_{{\mathbb{R}}^r}^r={\mathbb{R}}^r\times {\mathbb{R}}^r,$ the standard-coordinate identification. It is determined by the identity chart alone, so no choice is made in exhibiting it.

Pulling this identification back along $f$ gives a smooth bundle isomorphism $f^{\ast }T{\mathbb{R}}^r\cong f^{\ast }{\epsilon }_{{\mathbb{R}}^r}^r$ over $N$, and [F2] gives a canonical isomorphism $f^{\ast }{\epsilon }_{{\mathbb{R}}^r}^r\cong {\epsilon }_N^r$ carrying the pulled-back constant frame to the standard frame. Composing, $f^{\ast }T{\mathbb{R}}^r\cong f^{\ast }{\epsilon }_{{\mathbb{R}}^r}^r\cong {\epsilon }_N^r$ canonically. The only choice principle used is ${\mathrm{A}\mathrm{C}}_{\omega }$, inherited through [F1]; the pullback comparison of [F2] is choice-free, and the empty or disconnected case of $N$ is included since all maps displayed are evaluated fibrewise.