I am reading the first chapter of the Tu's book "Introduction to Manifolds" and I have found this (apparently innocent) excercise.
Let $X$ be the vector field $x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}$ and $f(x,y,z)$ the function $x^2+y^2+z^2$ on $\mathbb{R}^3$. Compute $Xf$.
In the previous sections a directional derivative $D_p$ is shown to belong to the tangent space $T_p(\mathbb{R}^n)$ as an $n$ fold vector i.e. with the same dimensions of $p$. In the excercise, instead, the tangent space "appears" to be $2$-dimensional while $f$ has $\mathbb{R}^3$ as domain.
I would be tempted to say that $Xf = 2x^2+2y^2$ i.e. the vector field $X$ transforms $f$ in a function with constant value w.r.t. $z$, or, also, that I can see $X$ as $x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}+0\frac{\partial}{\partial z}$ as if the field is living in a 2 dimensional subspace of a 3 dimensional tangent space... or.. am I loosing something here and the author meant something deeper?
Any suggestion welcome and sorry for the silly question.
I believe that Tu introduces the tangent space of $M$ at $p$ as the set of $\mathbb{R}-$derivations of $\mathscr{C}^\infty_p(M)$. A vector field over $U\subseteq M$ is then an element of $\operatorname{Der}_{\mathbb{R}}(\mathscr{C}^\infty(U)).$ That is, it is a $\mathbb{R}-$linear operator $\partial: \mathscr{C}^\infty(U)\to \mathscr{C}^\infty(U)$ satisfying the Leibniz rule $\partial(fg)=g\partial(f)+f\partial(g)$.
In our case, we have $\mathbb{R}^3=M=U$, and the derivation $\partial=X=x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}.$ Given a point $p\in U$ we can extract from this vector field a tangent vector, $X_p=x(p)\frac{\partial}{\partial x}|_p+y(p)\frac{\partial}{\partial y}|_p$. Indeed, pointwise $X_p$ lives in a two dimensional subspace of $T_p\mathbb{R}^3$ spanned by $\frac{\partial}{\partial x}|_p$ and $\frac{\partial}{\partial y}|_p$.
Applying $X$ to the function $f$, we get a function $Xf\in \mathscr{C}^\infty(\mathbb{R}^3)$ by the natural rule: $$ X(f)=x\cdot \frac{\partial f}{\partial x}+y\cdot \frac{\partial f}{\partial y}=2x^2+2y^2$$ as you wrote. The fact that this function does not depend on $z$ is a coincidence. For instance, if we chose $g=xyz$, then $$ X(g)=xyz+xyz=2xyz.$$