Let $X$, $Y$ and $Z$ be smooth manifolds and $$ X\xrightarrow{~f~}Y\xrightarrow{~g~}Z $$ be two composable embeddings. One can consider the normal bundle $\nu(f)\to X$ and $\nu(g)\to Y$ and the normal bundle $\nu(g\circ f)\to X$ of the composition.
Is $\nu(g\circ f)\cong \nu(f)\oplus f^*\nu(g)$ true?
Motivation: By definition we have $\nu(a)\oplus T_A\cong a^*(T_B)$ for $a\colon A\to B$ where $T$ denotes the tangent bundle. Hence $T_X\oplus \nu(g\circ f)\cong T_X\oplus \nu(f)\oplus f^*\nu(g)$ but we cannot remove the summand $T_X$ in the unstable situation.
Yes, this is true. You have three natural short exact sequences
$$0 \to T_Y \to g^{\ast} T_Z \to \nu(g) \to 0$$ $$0 \to T_X \to f^{\ast} T_Y \to \nu(f) \to 0$$ $$0 \to T_X \to f^{\ast} g^{\ast} T_Z \to \nu(gf) \to 0.$$
The map $T_X \to f^{\ast} g^{\ast} T_Z$ can be identified with the composite of the map $f^{\ast} T_Y \to f^{\ast} g^{\ast} T_Z$ coming from applying $f^{\ast}$ to the first short exact sequence and the map $T_X \to f^{\ast} T_Y$ coming from the second short exact sequence. This means that after choosing splittings, so that you get an isomorphism
$$T_X + \nu(gf) \cong T_X + \nu(f) + f^{\ast} \nu(g)$$
(here I'm writing $+$ for the direct sum) as you say, and this isomorphism respects the inclusion of $T_X$ on both sides, so you can cancel it (by quotienting by it).