Explaining problem in Gadea's "Analysis and Algebra on Differentiable Manifolds" I have a lot of trouble trying to explain to myself what the author did in problem 1.102 (the answer is in the link):

Let $TM$ be the tangent bundle over a differentiable manifold $M$. Let $\varphi: \mathbb R\times TM \to TM$ defined by $\varphi(t, X)=e^tX$.
   (i) Prove that $\varphi$ is a $1$-parameter group of transformations of $TM$.
   (ii) Calculate the vector field $Y$ on $TM$ associated to $\varphi$.
   (iii) Prove that $Y$ is invariant under $\varphi$.

I'm interested in the first point since maybe the rest continues from in.
The definition of $1$-parameter group (or flow) is given for maps that go from $M\times \mathbb R$ to $M$, where $M$ is a smooth manifold. Althought we know that the tangent bundle $TM$ gives a manifold structure to tangent vectors, applying straight the definition of flow doesn't seem like the way  the author took. $(1)$ I don't uderstand why does he prove that $\varphi_t \circ \varphi_s=\varphi_{t+s}$, I saw that he used that too in the problem before. $(2)$ Later when he is defining the chart in $TM$, I don't quite get the definition of $\Psi$. I thought the charts in $TM$ where given like this: if you take a chart of $M$, say $(U,\phi)$ then it induces the chart of the tangent bundle  like $(\pi^{-1}(U),\tilde \phi)$, with $\tilde\phi(X_p)=(\phi(p),\omega)$ and $\omega\in T_pM$. I don't know if my diagram is correct, but that's what I understand, and I don't see how did he got the so called $\Psi$ and $\tau$.

 A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\dd}{\partial}\renewcommand{\phi}{\varphi}$If $M$ is a smooth manifold, a one-parameter group on $M$ is usually defined to be a smooth homomorphism from $(\Reals, +)$, the additive group of real numbers, into the diffeomorphism group of $M$ under composition.
The condition $\phi_{t+s} = \phi_{t} \circ \phi_{s}$ is the homomorphism condition.
I didn't read the linked solution, but perhaps it helps first to look at $M = \Reals^{n}$, so $TM = T\Reals^{n} = \Reals^{n} \times \Reals^{n}$. Let $x$ denote base coordinates and $v$ denote fibre coordinates. If I understand the notation, the one-parameter group in question is $\phi(t, x, v) = (x, e^{t}v)$, homothetic scaling in the fibres. The vector field generating the flow is the Euler field
$$
\frac{\dd}{\dd t}\bigg|_{t=0} \phi(t, x, v) = (0, v),
$$
whose value at $(x, v) \in TM$ is $(0, v) \in T_{(x, v)}(TM)$.
Generally, if $x = (x_{1}, \dots, x_{n})$ are local coordinates on $U \subset M$, then the coordinate vector fields $\dd_{1}, \dots, \dd_{n}$ are a frame for $TU = TM|_{U}$. The "natural" coordinates on $TU$ are
$$
\tilde{\phi}\left(x, \sum_{j=1}^{n} v_{j} \frac{\dd}{\dd x_{j}}\right)
  = (x_{1}, \dots, x_{n}, v_{1}, \dots, v_{n}).
$$
(Your picture looks fine; the only potential missing detail is the relationship of the fibre coordinates $\omega$ with the coordinate vector fields.)
Now, the flow mapping is $\phi(t, x, v) = (x, e^{t}v)$, namely
$$
\phi(t, x_{1}, \dots, x_{n}, v_{1}, \dots, v_{n})
  = (x_{1}, \dots, x_{n}, e^{t}v_{1}, \dots, e^{t}v_{n}).
$$
Formally, everything looks exactly as if $M = \Reals^{n}$.
