$\newcommand{\Vec}{\mathbf{v}}\newcommand{\Pt}{\mathbf{p}}$This is an exercise in Barrett O'Neil's Elementary Differential Geometry 2nd Edition, p. 110 (Section 3.2)
If $T$ is a translation, show that for every tangent vector $\Vec$ the vector $T(\Vec)$ is parallel to $\Vec$.
My question is:
Isn't it the tangent map of $T$, not $T$ itself, which transform a tangent vector $\Vec$ at a point $\Pt$ to $T*(\Vec)$ at a point $T(\Pt)$?
(The star $*$ denotes tangent map in this textbook.)
I mean, I wonder whether it is possible to denote like $T(\Vec)$, which means that $T$ is directly applied to a tangent vector, not a point...
A translation looks like $T(\mathbf x) = \mathbf x + \mathbf c$ for some fixed vector $\mathbf c$ (I assume everything's in $\Bbb R^3$, but it doesn't matter).
Any mapping $f$ induces an action on tangent vectors at $p$ by taking a curve $\boldsymbol\gamma(t)$ with $\boldsymbol\gamma(0)=p$ and $\boldsymbol\gamma'(0)=\mathbf v$ and finding the tangent vector to the curve $f(\boldsymbol\gamma)$ at $t=0$. Note that this will be a tangent vector at $f(p)$.
In our case, using $f=T$, for any such curve $\boldsymbol\gamma$, we have $T(\boldsymbol\gamma(t)) = \boldsymbol\gamma(t)+\mathbf c$, so the derivative at $t=0$ is $\boldsymbol\gamma'(0) + \mathbf 0 = \mathbf v$. But we think of $\mathbf v$ as sitting (tangent) at $p$ and this new vector as sitting (tangent) at $T(p)$.
By the way, you should not write $T^*(\mathbf v)$. I do not know O'Neil's notation, but it certainly should not have an upper star.