On Wikipedia, it says
On every tangent bundle $TM$, considered as a manifold itself, one can define a canonical vector field $V : TM → TTM$ as the diagonal map on the tangent space at each point. This is possible because the tangent space of a vector space W is naturally a product, $TW \cong W \times W$, since the vector space itself is flat
https://en.wikipedia.org/wiki/Tangent_bundle
Does anyone know if there is a proof that $TW \cong W \times W$? Thanks!
For any point $p\in W$, any vector $v\in W$, one can define a path $\gamma:t\mapsto p + t v$: clearly $\gamma(0) =p$ and $\gamma'(0)=v$. Conversely, given a path $\gamma$, with $\gamma (0) =p$, $$ {\gamma(h) -\gamma (0)\over h }\in W,$$ for all (small, say) non-zero $h$. Hence, $\gamma'(0)\in W$. Therefore, we have the natural identification $TW = \{(p,v) \in W\times W \}$.
Addition
The following is in response to the comment below asking for motivation and a clarification about 'what is' $T_xW$.
I think most people secretly think of the tangent space $T_xW$ fairly concretely, a la vector calculus, much as above, as the space of derivatives $\gamma'(0)$ of curves $\gamma:\mathbb R \to W$ through the point $x=\gamma(0)$. This works, as is, if one knows that the manifold is sitting inside some kind of ambient Euclidean space, because then we can calculate and define a tangent vector $v$ by
$$ v= \gamma'(0) = \lim_{h\to 0} {\gamma(0+h) -\gamma(0)\over h}.$$ That is, one can do this because the above difference, quotient, and limit make sense. For general manifolds $W$, one has to go through some kind formalism for any definition or calculation, but the result has to be consistent with above. I would say that the sages have done this by replacing the frame work provided by an'ambient space' with one that uses nice test functions $f \colon W \to \mathbb R$: the composition $f\circ \gamma$ is a function from $\mathbb R$ to $\mathbb R$, so one can again understand $$ {d\over dt}\Bigg\rvert_{t=0}\, f \circ \gamma = \lim_{h\to 0} {(f\circ\gamma)\,(0+h) -(f\circ\gamma)\,(0)\over h}.$$ At this point one identifies a tangent vector $v$ with the set of $\gamma$ for which the preceding gives the same result for any fixed test function $f$, i.e., one defines $v$ as functional (function on some set of functions - or more correctly, on some set of equivalence classes of functions) by the formula
$$ v(f)= {d\over dt}\Bigg\rvert_{t=0}\, f \circ \gamma,$$ for some equivalence class of $\gamma$. This matches up with the vector calculus version, because of the chain rule.
How one does this - or how the sages do this - formally and rigorously depends on the approach to define a manifold - i.e. by charts and atlases, or as a locally ringed space - and what equivalent choice of definition of 'functional' one uses (derivation, or as an element of the dual $m/m^2$, where $m$ is the maximal ideal of functions - or equivalence classes of functions - vanishing at $x$)... One has to show that given the 'functional' $v$ - i.e., 'the' tangent vector $v$ - that there is a corresponding curve $\gamma$ - but that is 'clear' because of charts, say, and because one can do so in $\mathbb R^n$. For a more complete and (somewhat official) reference, see the "Formal Definitions" section of https://en.wikipedia.org/wiki/Tangent_space.
N.B. I hope that this addition motivates and explains the first part of the answer - i.e., given a point $x\in W$, and $v\in W$ ($W$ is now not the arbitrary manifold, but the vector space), there is a corresponding curve $\gamma$ such that $\gamma(0)= x$ and $\gamma'(0)=v$, and vice versa; this identification corresponds to the construction above. Hence we can identify $T_xW = W$, and $TW= W\times W$.