Here are explicit local isomorphisms which you can easily check patch together. Notice that the first inverse is given in vakil's notes above the exercise, and the rest are gotten the same way. Also notice that you only need the first four of these to define the isomorphism since $\mathbb{P}^1$ (resp the conic $\mathcal{C}$) is covered by the domains of those isomorphisms.
$$k[m]_{m^2+1} \cong \big(k[x,y]/(x^2+y^2-1)\big)_{x-1},$$
$$ m\mapsto y/(x-1),~~~~~~~~ (x,y)\mapsto (\frac{m^2-1}{m^2+1}, \frac{-2m}{m^2+1}). $$
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$$k[n]_{n^2+1}\cong \big(k[x,y]/(x^2+y^2-1)\big)_{x+1},$$
$$ n\mapsto -y/(x+1) ,~~~~~~~~~(x,y)\mapsto (\frac{1-n^2}{n^2+1}, \frac{-2n}{n^2+1}) .$$
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$$k[m]_{m^2-1}\cong \big(k[y,z]/(y^2-z^2+1)\big)_{z-1},$$
$$ m\mapsto y/(1-z),~~~~~~~~~ (z,y)\mapsto (\frac{m^2+1}{m^2-1}, \frac{-2m}{m^2-1}). $$
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$$k[n]_{1-n^2}\cong \big(k[y,z]/(y^2-z^2+1)\big)_{z+1},$$
$$ n\mapsto -y/(1+z), ~~~~~~~~~~(z,y)\mapsto (\frac{n^2+1}{1-n^2}, \frac{-2n}{1-n^2}) $$
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$$k[m]_m\cong \big(k[x,z]/(x^2-z^2+1)\big)_{x-z},$$
$$ m\mapsto 1/(x-z),~~~~~~~~~ (z,x)\mapsto (\frac{-m^2-1}{2m}, \frac{1-m^2}{2m}). $$
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$$k[n]_n\cong \big(k[x,z]/(x^2-z^2+1)\big)_{x+z},$$
$$ n\mapsto -1/(x+z), ~~~~~~~~~~~ (z,x)\mapsto (\frac{-n^2-1}{2n}, \frac{1-n^2}{2n}).$$