Why is the relative tangent bundle called so and why is a bundle of tangent vectors to the fibre? Let $X$ and $Y$ be smooth manifolds and $\varphi:X\to Y$ be a submersion. Let $T_X$ and $T_Y$ be the tangent bundles over $X$ and $Y$ respectively. The relative tangent bundle $T_{X/Y}$ is defined as the kernel of the map $T_\varphi:T_X\to\varphi^*T_Y$.
Question 1: Why is this bundle called "relative" tangent bundle?
Question 2: Why can it be thought of as the bundle of tangent vectors that are tangent to fibres of $\varphi$?
What we have is a short exact sequence of vector bundles over $X$ - $$0\to T_{X/Y}\to T_X\xrightarrow{T_\varphi} \varphi^*T_Y\to 0$$ So the over a point $x\in X$ we have the SES of vector spaces $$0\to T_{X/Y,x}\to T_{X,x}\xrightarrow{T_{\varphi,x}} T_{Y,\varphi(x)}\to 0$$
Where $T_{X/Y,x}=\{u\in T_{X,x}\ |\ T_{\varphi,x}(u)=0\}$. But it is not clear to me why such $u$ are precisely the tangents to the fibres?
Thank you.
 A: a) If $\varphi:X\to Y$ is a submersion and $y=\varphi(x)$, then the fibre $X_y=\varphi^{-1}(y)\subset X$ is a submanifold whose tangent space at $x$ is the linear subpace $T_x(X_y)\subset T_x(X)$ defined by $$T_x(X_y)=\operatorname {Ker}[T_x(X)\xrightarrow {D{\varphi }_x }T_{\varphi(x)}(Y)] $$ Hence we have for each $x\in X$ an exact sequence of vector spaces $$   0\to T_x(X_y)\to T_x(X) \xrightarrow{D\varphi_x} T_{\varphi(x)}(Y) \to  0 \quad (\bigstar)_x                 $$ This has nothing yet to do with vector bundles and follows from the basic properties of submersions: see Lee, Proposition 5.38, page 117 whose notations I have adopted for ease of reference, although your notation too is  beautifully functorial!
b) The collection of all those subspaces $T_x(X_y)\subset T_x(X)$ for varying $x\in X$ forms a sub-vector bundle $T(X/Y)=T_{\operatorname {vert}}(X)\subset T(X)$ called the vertical tangent bundle of $X$ with respect to $\varphi$ (a more adequate terminology than relative tangent bundle).
The exact sequence
$$0\to T_{\operatorname {vert}}(X)\to T(X) \xrightarrow{T(\varphi)} \varphi^* T(Y)\to 0 \quad ( \bigstar)$$ of vector bundles on $X$ is thus  obtained by the globalization of  the exact sequences of vector spaces $(\bigstar)_x$.
Note carefully that there does not exist a   canonical horizontal bundle $T_{\operatorname {hor}}(X)\subset T(X) $ such that $$T(X)=T_{\operatorname {vert}}(X)\oplus T_{\operatorname {hor}}(X)$$ In other words our exact sequence of vector bundles  $ (\bigstar )$ has no canonical splitting.
