On a Riemannian manifold $M$ one can identify 1-forms and vector fields by $$ \alpha(p) = \langle X(p),\cdot\rangle_p $$
Since we can perform a pullback on both 1-forms and vector fields I expected that for a diffeomorphism $\phi:M\to M$ we have $$\phi^*\alpha(p)=\langle\phi^*X(p),\cdot\rangle_p$$ However, since $$\phi^*\alpha(p)=\alpha(\phi(p)\textrm D \phi^{-1}(\phi(p)) = \langle X(\phi(p)),\textrm D\phi^{-1}(\phi(p))\cdot\rangle_{\phi(p)}$$ and $$\langle\phi^*X(p),\cdot\rangle_p=\langle\textrm D\phi^{-1}(\phi(p))X(\phi(p)),\cdot\rangle_p$$ my guess seems to be wrong. Is there anything wrong with my calculation?
Are you sure about your definition of the pullback of a one-form? My definition would be $$ (\phi^*\alpha)_p(Z_p) = \alpha_{\phi(p)}(D \phi_p (Z_p)) = \langle X_{\phi(p)}, D \phi_p (Z_p) \rangle_{\phi(p)}, $$ your expressions with $D\phi^{-1}$ don't make much sense (note that $D\phi^{-1}_{\phi(p)}$ maps into $T_pM$, so you can't take the scalar product at $\phi(p)$ on the r.h.s.) ...
Now, in order to make your identifications (which depend on the metric!) compatilbe with taking the pullback by a diffeomorphism $\phi$, you have to assume that $\phi$ is not only a diffeomorphism but also an isometry, i.e. $$ \langle \cdot , \cdot \rangle_p = \langle D\phi(p) \cdot , D\phi(p) \cdot \rangle_{\phi(p)} $$ Just plug in this into the calculations and you'll be fine :)