I'm reading about isometric immersions in the Do Carmo's Riemannian Geometry and there he states that if $f: M^n \longrightarrow \overline{M}^{n+m=k}$ is an immersion and $\overline{M}$ endowed with an inner product, then we can splits $T_p \overline{M}$ into the direct sum
$$T_p \overline{M} = T_p M \oplus \left( T_p M \right)^{\perp},$$
where $\left( T_p M \right)^{\perp}$ is the orthogonal complement of $T_p M$ in $T_p \overline{M}$.
My doubt is why is valid? I know from Linear Algebra that if $V$ is a vectorial space with dimension $n$ and $U \subset V$ is a subspace of $V$, then it's possible write $V = U \oplus U^{\perp}$, but $T_p M$ is not necessarily a subspace of $T_p \overline{M}$, so why we can decompose $T_p \overline{M}$ in a direct sum of $T_p M$ and $\left( T_p M \right)^{\perp}$?
Thanks in advance!
$T_p M$ is naturally identified with a subspace of $T_{f(p)} \overline M$ via the map $Tf\colon T_p M \to T_{f(p)} \overline M$. Here $Tf$ is injective on $T_p M$ because $f$ is required to be an immersion.