2 corrected an error with \chi(M)

I assume that your surface is closed. Suppose you have a fixed vector bundle $\xi$ over $S^2$ (no matter which one). You have an oriented surface $M$ embedded in $\mathbb R^3$, which defines the Gauss map $\nu:M\to S^2$, which defines the vector bundle $\nu^*\xi$ on $M$. You want to know whether $\nu^*\xi$ depends on the embedding.

No it does not. Indeed, $\deg\nu=\chi(M)$ as you mentioned\deg\nu=\chi(M)/2$regardless of the embedding. Two maps$f_1,f_2:M\to S^2$having the same degree are homotopic. And homotopic maps induce the same bundle. Concerning the Chern class, we have $c_1(\nu^*\xi)=\nu^*(c_1(\xi))$ by definition. So, if you identify the top cohomology with integers, then $c_1(\nu^*\xi)$is (the same number as)$\deg(\nu) c_1(\xi)=\chi(M)c_1(\xi)$c_1(\xi)=\frac12\chi(M)c_1(\xi)$.

1

I assume that your surface is closed. Suppose you have a fixed vector bundle $\xi$ over $S^2$ (no matter which one). You have an oriented surface $M$ embedded in $\mathbb R^3$, which defines the Gauss map $\nu:M\to S^2$, which defines the vector bundle $\nu^*\xi$ on $M$. You want to know whether $\nu^*\xi$ depends on the embedding.

No it does not. Indeed, $\deg\nu=\chi(M)$ as you mentioned. Two maps $f_1,f_2:M\to S^2$ having the same degree are homotopic. And homotopic maps induce the same bundle.

Concerning the Chern class, we have $c_1(\nu^*\xi)=\nu^*(c_1(\xi))$ by definition. So, if you identify the top cohomology with integers, then $c_1(\nu^*\xi)$ is (the same number as) $\deg(\nu) c_1(\xi)=\chi(M)c_1(\xi)$.