(in fact Whitney theorem says that we can embed $M$ to ${R}^{2d}$)
Let $N\to M$ be the normal bundle of the embedding $M\subset {R}^k$. We know that $N\oplus TM={R}^k$ is a trivial bundle on $M$. You want to know why the bundles $N$ and $f^*N$ are isomorphic. Unfortunately I can't prove this. However they are stably isomorphic: there in $n$ s.t. $N\oplus\mathbb{R}^n$ and $f^*N\oplus {R}^n$ are isomorphic. For $n$ we can take $k$. Namely, $f^*TM\cong TM$, hence $R^k\oplus N\cong f^*R^k\oplus N\cong f^*N\oplus f^*TM\oplus N\cong f^*N\oplus TM\oplus N\cong f^* N \oplus R^k$.
Altogether it means that if you embed $M$ into $R^k$ and $R^k$ standardly into $R^{2k}$ then you have your tubular neighbourhood isomorphism for the resulting embedding $M\to R^{2k}$. Perhaps it's sufficient for your purposes.