Things are much nicer i.e. more general and canonical than that!

a) Let $\pi:N\to M$ be a submersion of manifolds. From it you obtain the exact sequence of vector bundles on $N$: $$ 0\to T^{vert}(N) \to T(N) \stackrel {d\pi}{\to} \pi^{-1}T(M) \to 0 \quad (*) $$ Profound, eh? Not at all!

This is just a fancy way of looking at the differential of the map $\pi: N\to M.$

In order that both the tangent bundles to $N$ and to $M$ live on $N$, you have to pull back $T(M)$ to $N$: that is why we have $\pi^{-1} T(M)$ on the right.

The kernel is the set of vertical tangent vectors to $N$, those that lie along the level lines $N[n]\stackrel {def}{=}\pi^{-1}(\pi (n))$ of $\pi$.

In other words at a point $n\in N$ the fibre of $T^{vert}(N)$ is $$T^{vert}_n(N)=T_n(N[n]) \quad (**)$$

b) Consider now a vector bundle $\pi:V\to M$ on $M$.

You then have the canonical exact sequence of vector bundles on $V\;$ (yes, vector bundles on a vector bundle!) :

$$ 0\to T^{vert}(V) \to T(V) \stackrel {d\pi}{\to} \pi^{-1}T(M) \to 0 \quad (***)$$

In this new set-up you have the interesting identification $ T^{vert}(V)=\pi^{-1}(V)$.

This boils down to the fact that the tangent space at any point $e\in E$ of a vector space $E$ is that vector space itself: $T_e(E)=E.$

Hence $(***)$ becomes $$ 0\to \pi^{-1}(V) \to T(V) \stackrel {d\pi}{\to} \pi^{-1}T(M) \to 0 \quad (****) $$

c) Finally, if you restrict this last exact sequence (****) to the zero section of $V$, identified to $M$, you get

$$ 0\to V \to T(V)|M \stackrel {d\pi}{\to} T(M) \to 0 \quad \quad (*****) $$

d) Although this is not very helpful, you can if the manifold $M$ is paracompact (see here) *non-canonically* split the exact sequence (*****) and obtain the isomorphism $$ T(V)|M \cong V \oplus T(M)$$
At a point $(m,0)\in M\subset V$ this gives the decomposition $$ T_{(m,0)}(V) \cong V_m \oplus T_m(M)$$
and for $V=T^*(M)$ this is what you were looking for.