Normal bundle in tangent bundle

Actually a much more general result is true: given a manifold $M$ and an arbitrary vector bundle $E\stackrel {\pi}{\to} M$ on $M$, the normal vector bundle $N_E(M)$ of $M$ (identified with the zero section of $E$) in $E$ is isomorphic to $E$.
In other words we have a direct sum decomposition of vector bundles on $M$:

$$T(E)|M =T(M)\oplus E $$

The key to understanding this is to consider the case where $M$ is reduced to a point: it is then the result which says that the tangent space at the origin $T_0E$ of a vector space $E$ can be identified with the vector space $E$ itself.
Indeed, if tangent vectors are defined (say) as derivations, then the vector $v\in E$ is identified with the derivation $\partial _v|_0$ which sends the function $f$ to its directional derivative $$\partial _vf(0)=\lim_{t\to 0} \frac {f(tv)-f(0)}{t}$$ Edit
It might be of interest to notice that associated to any vector bundle $E\stackrel {\pi}{\to} M$ we have a canonical exact sequence of bundles on $\textbf E$ : $$0\to\pi^*(E)\to T(E)\to \pi^*(T(M))\to 0 $$ The vector bundle $\pi^*(E)=:T_{vert}(E)$ is the subvector bundle of $T(E)$ consisting of vectors tangent to the fibers of $\pi$.
Restricting this exact sequence to the zero section of the bundle identified with $M$ we get the canonical exact sequence of bundles on $\textbf M$: $$0\to E\to T(E)\vert M\to T(M)\to 0$$ This exact sequence can non canonically be split and we obtain the already mentioned decomposition $T(E)|M =T(M)\oplus E$ .