Given $v=v^1\partial_\theta +v^2\partial_\varphi$ a vector in $TS^2$ in polar coordinates, I would like rotate it of $\pi/2$ degrees. Looking at $S^2$ into $R^3$, the rotation is given by the cross product with the normal vector at the surface: $\mathbf{n}\times v$.
What is the equivalence in spheric coordinates? Is there a way to take advantage of the Riemannian metric?
On
You can do it with the help of Levi-Civita tensor.
Define $\epsilon_{ij}\mathbf{n}=\mathbf e_i \times\mathbf e_j$ (with $\mathbf n=\frac{\mathbf e_\theta \times\mathbf e_\phi}{|\mathbf e_\theta \times\mathbf e_\phi|}$). In polar coordinates ($\theta\in[-\pi/2, \pi/2]$, $\phi\in[0,2\pi)$) you can explicitly write: $$ \epsilon_{\theta\theta}=\epsilon_{\phi\phi}=0,\qquad \epsilon_{\theta\phi}=-\epsilon_{\phi\theta}=\cos\theta. $$
Then, to rotate a vector on angle $\psi$, one would use this formula: $$ \bar v^i=v^i\cos\psi + \epsilon^{\ \ i}_{j}v^j\sin\psi $$
In your particular case: $\bar v=-\cfrac{v^2}{\cos\theta}\partial_\theta+v^1\cos(\theta)\partial_\phi$
Assuming you're using the non-American spherical coordinates ($\phi\in (0,\pi)$, $\theta\in (0,2\pi)$), $\partial_\theta$ is a unit vector and $\partial_\phi$ has length $\sin\theta$, and they form a right-handed coordinate system. Thus $\mathbf n\times\partial_\theta = \frac1{\sin\theta}\partial_\phi$ and $\mathbf n\times\partial_\phi = -\sin\theta\,\partial_\theta$. Now just expand $\mathbf n\times v$.