In Jost's Riemannian Geometry and Geometric Analysis (6th Edition), lemma 4.3.4 claims that $d=\eta^i\wedge \nabla_{e_i}$, where $\{e_i\}$ is a local orthonormal frame field and $\{\eta^i\}$ is its dual (so that $\eta^j(e_i)=\delta^j_i$). I understand that we can prove this by using normal coordinate on each point as the formula is independent to the choice of local frame(doesn't even need to be orthonormal, as pointed out in the Remark in the book).
My question is, for any coordinate since $\{\partial_{x_i}\}$ also gives a local frame with dual frame $\{dx_i\}$. Then the formula indicate that $$d=\epsilon(dx_i) \nabla_{\partial_{x_i}},$$ which is absurd because in local coordinate we have $$d=\epsilon(dx_i)\partial_{x_i}.$$ I must misunderstand something in the material. Thanks in advance for clearing up the idea!
I assume that you are using Levi-Civita connection. Suppose that $\omega = \omega_k dx^k$ is any 1-form then $\nabla_{\partial_i}\omega_j = \partial_i\omega_j - \Gamma_{ij}^k\omega_k$. Therefore, \begin{align} d(\omega_kdx^k) &= \epsilon(dx^i)\nabla_{\partial_i}(\omega_kdx^k) = dx^i\wedge((\partial_i\omega_k)dx^k - \Gamma_{ik}^j\omega_jdx^k)\\ &= dx^i\wedge(\partial_i\omega_k)dx^k - \Gamma_{ik}^jdx^i\wedge dx^k \omega_j \end{align} where there are implies summation over repeated indices. Since $\Gamma_{ik}^j = \Gamma_{ki}^j$, the second term vanishes, and we have \begin{equation} d(\omega_kdx^k) = dx^i\wedge(\partial_i\omega_k)dx^k = \epsilon(dx^i)\partial_i\omega \end{equation} as required.