I was asking myself if the tangent bundle of a trivial bundle $\mathcal{P}=M\times V$ with fiber $\pi: \mathcal{P}\to M$ (actually it would be a principal trivial bundle, but I think it doesn't matter for the question), where $V$ is a vector space (where a certain group $G$ acts), has same fibers of the tangent bundle over $M$, so if this holds:
$$T_p\mathcal{P}=T_xM,$$
where $p\in \mathcal{P}$ such that $\pi(p)=x\in M$.$$$$ And therefore $$\bigcup_{p\in \mathcal P} \{p\}\times T_p \mathcal P = V \times \bigcup_{x\in M}\{x\}\times T_x M .$$
*edit: of course (according to last equation too) $T_p\mathcal{P}=T_xM\oplus V,$
In general for a bundle of smooth manifolds $N\to E \to M$ with projection $\pi$, locally $E$ looks like $U_x\times N$ for $U_x$ an neighbourhood of $x\in M$, and wrt to this chart the tangent space at a point $(x,y)$ looks like $T_{(x,y)}E\cong T_xM\times T_yN$. There is even a short exact sequence of bundles
$$ ker(D\pi) \to TE \to TM $$
and choice of connection on $E$ corresponds to a choice of splitting for this sequence, and hence a direct sum decomposition of $TE$ into "vertical" and "horizontal" vectors.
Now let $M^n$ be a smooth manifold, $TM$ its tangent bundle. At any point $(x, v)$ in $TM$ the tangent bundle looks like $T_{(x,v)}\cong T_pM\times T_vT_pM\cong \mathbb{R}^{2n}$, so the fibres of $TTM\to TM$ have twice the dimension as the fibres of $TM \to M$. In fact if we assume as in your question that our tangent bundle is trivial, then there is a global isomorphism
$$ T(TM) \cong TM \times \mathbb{R}^{2n} $$