As we know, the topological $K$-group $K_\mathbb{R}(S^2)=\mathbb{Z}\bigoplus\mathbb{Z}/2$. Which bundle represents the $\mathbb{Z}/2$ part？
I know that the tangent bundle of S$^2$ is not trival by hairy ball theorem, maybe they have some relation, but how to check it has order 2?
The tangent bundle of $S^2$ is stably trivial, so it's already zero in K-theory.
To answer this, first let's actually just figure out what all the real vector bundles over $S^2$ are.
Bundles over $S^2$ are automatically orientable, and oriented bundles are in bijection with $[S^2,BSO(n)] = \pi_1(SO(n))$, which is $\Bbb Z$ for $n=2$ and $\Bbb Z/2$ for $n>2$. What are the actual bundles here? For $n>2$, we learn that there are only two bundles up to isomorphism. One is trivial, and the other has nontrivial $w_2$ - you can obtain it by taking a 2-plane bundle with $w_2\neq 0$ and summing with a trivial bundle of rank $n-2$.
What about $n=1$? These circle bundles can be distinguished by their Euler class, which is an integer. As unoriented bundles, they're isomorphic iff $|\chi_1| = |\chi_2|$. The one with Euler class 1 is the tautological bundle, and has $w_2 \neq 0$. And stably (after summing with the trivial rank 1 bundle) they're isomorphic iff they have the same $w_2$, just by our previous calculation of the rank 3 bundles. Note that the bundle with $\chi = 2$ is the tangent bundle.
So your K-theory is generatedy by the trivial rank one bundle $\epsilon$ (that's the $(1,0)$ generator) and the tautological bundle $\xi$ over $\Bbb P^1$ (which corresponds to $(1,1)$). The vector bundle of order two is actually a virtual vector bundle $\xi-\epsilon$. (Given that the dimension provides a homomorphism $K(S^2) \to \Bbb Z$, any torsion must live in dimension 0, so must be a virtual bundle.)