I am trying to solve a problem from John M. Lee's Riemannian Manifolds. The goal is to show that if we have a vector field $X = X^i E_i$ in terms of some local frame $\{E_i\}$ on $U\subset M$, then we can write the divergence in terms of covariant derivatives: $$\text{div}(X) = X^i_{;i}$$
Where (in a coordinate frame/coframe): $$\nabla X = X^i_{;j}\partial_i\otimes dx^j = \left(\partial_j X^i + X^k\Gamma_{jk}^i\right)\partial_i\otimes dx^j$$
The hint says to "show that it suffices to prove the formula at the origin in normal coordinates." Let $p\in U$ be given. Since $\{E_i\}$ is a local frame, $\{E_i|_p\}$ forms a basis of $T_pM$. Using the Gram-Schmidt algorithm, we can construct an orthonormal basis $\{F_i|_p\}$. With the isomorphism $F : \Bbb R^n\to T_pM$ together with the exponential map, $\varphi = F^{-1} \circ \exp_p^{-1} : \mathcal U \to \Bbb R^n$ gives us normal coordinates centered at $p$ with $\mathcal U \subset U$.
Now using the coordinate expression for divergence along with properties of normal coordinates: \begin{align*} \text{div}(X)|_p &= \left(\frac{1}{\sqrt{\text{det}(g_{ij})}}\sum_{k=1}^n\frac{\partial X^k\sqrt{\text{det}(g_{ij})}}{\partial x^k}\right)\Bigg|_p\\ &= \sum_{k=1}^n\frac{\partial X^k}{dx^k}\bigg|_p\\ &= X^k_{;k}(p) - X^\ell\Gamma_{k\ell}^k(p)\\ &= X^k_{;k}(p) \end{align*}
What I'm having a difficult time understanding is the "suffices" part of the hint. How do I extend that $\text{div}(X)=X^i_{;i}$ in normal coordinates to an arbitrary local frame? Especially since the definition of $X^i_{;j}$ includes a partial derivative, which need not be defined on an arbitrary frame?
A similar question is asked here: Divergence as the trace of total covariant derivative?
But I don't think it answers my question. That $X^i_{;i} = \text{tr}(\nabla X)$ gives a reason for frame-independence, but I don't see how it relates to the calculation in normal coordinates.
I'll add another possible solution without using normal coordinates.
In general for a vector space $V$ and $T\in End(V)$, $tr(T)=f_i\circ T(e^i)$ where $\{e_i\}_i$ is a basis for $V$ and $\{f_i\}_i$ is its dual basis. Since $$\text{div}(X)_p=tr\nabla_\bullet X_p=tr(Y_p\mapsto \nabla_{Y_p}X_p),$$ then we have $$\text{div}(X)_p=f^i(\nabla_iX_p)=f^i(E_i(X^j)E_j+X^j\nabla_iE_j)=E_i(X^i)+X^k\Gamma_{ik}^i=X^i_{;i}.$$ In normal coordinates this corresponds to one of your equations: $\text{div}(X)\mid_p=\partial_iX^i\mid_p$, since in normal coordinates centered at $p$ the Christoffel symbols are zero in $p$.