Consider a compact submanifold $X\hookrightarrow Y$ and its normal bundle $N\to X$: $$0\to TX\to TY|_X \to N \to 0$$
There is a diffeomorphism which identifies $N$ with a tubular neighborhood $\tilde N\subset Y$ of $X$.
Now move to the tangent spaces. It can be shown that $T\tilde N\subset TY$ is a tubular neighborhood of $TX\hookrightarrow TY$. Hence we can identify the total space $TN$ (which is the tangent space of the normal bundle) with the normal bundle of $TX$ in $TY$:
$$0\to TTX \to TTY|_{TX} \to TN \to 0$$
I would like to prove that $\pi^*N\oplus \pi^*N\simeq TN$, where $\pi\colon TX\to X$ is the projection. (And hence $TN$ can be given a structure of a complex vector bundle over $TX$).
This "simple verification is left to the reader" in M.F. Atiyah, I. Singer, Index of Elliptic Operators: I, page 497.
W can use the fact that $\pi$ is deformation retraction to get an exact sequence
$$0\to \pi^*TX \to \pi^* TY|_X \to \pi^*N \to 0$$
which we sum up with itself
$$0\to \pi^*TX \oplus \pi^*TX \to \pi^* TY|_X \oplus \pi^* TY|_X \to \pi^*N \oplus \pi^*N \to 0$$
Now using
$$TTX\simeq \pi^*TX \oplus \pi^*TX, \quad TTY|_{TX} \simeq \pi^* TY|_X \oplus \pi^* TY|_X$$
we get $$\pi^*N \oplus \pi^*N\simeq TN.$$
I wonder if the above sketch is right or if you have seen a similar topic explained somewhere. (In two books on the index theorem I know this lemma is very briefly commented).