**Edit:** According to interesting comment of Thomas Rot to the previous version of the question, we revise the question as follows:


First note that if a manifold $M$ is a parallelizable manifold , then it gets a natural Riemannian metric which is independent of the base point $x\in M$.In fact $TM \simeq M \times \mathbb{R}^n$ enable us to carry the standard Euclidean inner product to each fiber of $TM$. In the following question we apply this obvious fact to $M=TS^n$ as follows:


It is well known that the tangent bundle $TS^n$ of $S^n$ is a parallelizable manifold, then once we fix a trivialization for its tangent bundle(The tangent bundle of $TS^n$), $TS^n$ gets s a natural Riemannian metric, the Euclidean one along each fiber of its tangent bundle.
Now we can restrict this metric on $T S^n$ to the zero section $S^n$.



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> Is there a trivialization of $TS^n$ for which the corresponding restricted metric coincide to the standard metric of $S^n$?
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Can the standard Riemannian metric of $S^n$ be realized as the restriction of certain metric on $T S^n$?
You describe only a fiber metric on $TS^n$, an inner product on each fiber, but not a Riemannian metric on the total space. If you consider the metric induced from the embedding $Ti:TS^n \to T\mathbb R^{n+1} = \mathbb R^{n+1}\times \mathbb R^{n+1}$ (where $i:S^n\to \mathbb R^{n+1}$), then the answer is yes.


Further explanation:
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$i:S^n\subset\mathbb R^{n+1}$ is the standard embedding. The image of the tangent mapping $Ti$ consists of all $(x,\xi)\in \mathbb R^{n+1}\times \mathbb R^{n+1}$ such that $|x|=1$ and $\langle x,\xi\rangle = 0$. The Riemannian metric induced on $TS^n$ from the Euclidean metric on $\mathbb R^{n+1}\times \mathbb R^{n+1}$ is Euclidean on each fiber $T\_xS^n$ (fix $x$), and the metric on the zero section is the usual one (set $\xi=0$).


