Calculating the stiefel whitney class of Tangent Bundle of projective space

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I was reading the proof of the fact that whitney sum of the tangent bundle of $RP^n$ with the trivial bundle is isomorphic to whitney sum of $n+1$ tautological line bundles on $RP^n$ from Hatcher's Vector Bundles and K theory.

However I am confused at this particular point in which he says

In the normal bundle $NS^n $ the identification $(x, v) ∼ (−x, −v)$ can be written as $(x, tx) ∼ (−x, t(−x)) $. This identification yields the product bundle $RP^n × R$ since the section $x \to (−x, −x)$ is well-defined in the quotient. Now let us consider the identification $(x, v) ∼ (−x, −v)$ in $S^n × R^{n+1}$ . This identification respects the co- ordinate factors of $R^{n+1}$ , so the quotient is the direct sum of $n + 1$ copies of the line bundle $E$ over $RP^n$ obtained by making the identifications $(x, t) ∼ (−x, −t)$ in $S^n × R$ . The claim is that E is just the canonical line bundle over $RP^n $. To see this, let us identify $S^n × R $ with $NS^n$by the isomorphism $(x, t)\to(x, tx)$ ,hence $(−x, −t)\to ((−x, (−t)(−x)) = (−x, tx)$ . Thus we have the identification$ (x, tx) ∼ (−x, tx) $in $NS^n$ . The quotient is the canonical line bundle over $RP^n$ since the identifications $x ∼ −x$ in the first coordinate give lines through the origin in $R^{n+1}$ , and in the second coordinate there are no identifications so we have well-defined vectors $tx$ in these lines.

So it seems like he is proving that $(x,v) \sim (-x,-v)$ is trivial in the first three lines and later he seems to be proving that it is the tautological line bundle which is non trivial. Also I could not find this same proof anywhere else(all had the proof from the book characteristic classes by Milnor-Stasheff). If anyone can point out where I am going wrong or give some other reference it would be great. Thanks