Consider a 2-manifold $(M,g)$ isometric to $(\mathbb{R}^2,\bar{g})$, where $\bar{g}$ is the Euclidean metric, and let $F:\mathbb{R}^2 \to M$ be an isometry. Let $K_0 \subset \mathbb{R}^2$ and $K \subset M$ be two compact subsets (actually, I consider geodesic triangles, if that matters), such that $K = F(K_0).$
I consider two real-valued functions $f:K \to \mathbb{R}$ and $f_0 := f \circ F:K_0\to \mathbb{R}$, defined on $K$ and $K_0$ respectively, and I'm interested in the leading error term in Taylor's linear expansion of each function, which involves the Hessian.
On $(\mathbb{R}^2,\bar{g})$ with coordinates $y^i$, the Christoffel symbols vanish and the coefficients of the covariant Hessian (defined e.g. in Lee's Intro to Riemannian manifolds, Example 4.22) of $f_0$ are the usual Hessian matrix: $$(H_{f_0})_{ij} = \frac{\partial^2 f_0}{\partial y^i\partial y^j}.$$
On $(M,g)$ with coordinates $x^i$, we have: $$(H_{f})_{ij} = \frac{\partial^2 f}{\partial x^i\partial x^j} - \Gamma^k_{ij}\frac{\partial f}{\partial x^k}.$$
I am trying to recover the pullback of $H_f$ on $(\mathbb{R}^2,\bar{g})$ from the Hessian of the composition $f \circ F$ as follows. Since $F$ is an isometry, it pulls back the metric $g$ to the Euclidean metric, thus in components: \begin{equation} (F^*g)_{ij} = g_{kl} \frac{\partial F^k}{\partial y^i}\frac{\partial F^l}{\partial y^j} = \left(\frac{\partial F^i}{\partial y^k}\right)^T g_{kl} \frac{\partial F^l}{\partial y^j} = \bar{g}_{ij} = \delta_{ij}. \end{equation} This is satisfied when the Jacobian matrix of $F$ is of the form: \begin{equation} \tag{1} \frac{\partial F^i}{\partial y^j} = g_{ik}^{-1/2}O_{kj}, \end{equation} for some orthogonal matrix $O$, where the negative square root of $g_{ij}$ is defined by $g^{-1}_{mn} := g_{mp}^{-1/2}g_{pn}^{-1/2}$ (or alternatively using the negative square root of its eigenvalues as I wrote in this other question: Relation between square root of matrix representation of Riemannian metric and Christoffel symbols). Does it make sense to characterize the Jacobian of $F$ this way? I never see any mention of this negative square root matrix in textbooks or other questions on this forum.
Then, the Hessian of $f_0 = f \circ F$ at $p \in K$ writes: $$ (H_{f_0}(p))_{ij} = \frac{\partial^2 (f \circ F)}{\partial y^i \partial y^j}(p) = \frac{\partial^2 f}{\partial x^k\partial x^l} \biggr\vert_{F(p)}\frac{\partial F^k}{\partial y^i}(p)\frac{\partial F^l}{\partial y^j}(p) ~~\fbox{$+ \frac{\partial f}{\partial x^k}\biggr\vert_{F(p)}\frac{\partial^2 F^k}{\partial y^i \partial y^j}(p)$}. $$ On the other hand, the components of the pullback of $H_f$ at $p$ are: $$ (F^*H_f(p))_{ij} = \frac{\partial^2 f}{\partial x^k\partial x^l}\biggr\vert_{F(p)} \frac{\partial F^k}{\partial y^i}(p) \frac{\partial F^l}{\partial y^j}(p) ~\fbox{$ -\Gamma^p_{kl}\frac{\partial f}{\partial x^p}\biggr\vert_{F(p)}\frac{\partial F^k}{\partial y^i}(p) \frac{\partial F^l}{\partial y^j}(p)$}. $$ Are the boxed terms the same? Or maybe my understanding is wrong, and $H_{f_0}(p)$ and $F^*H_f(p)$ are not the same quantity? Replacing the Jacobian matrix of $F$ by the expression (1) and taking $O=I$ (the identity matrix) as orthonormal matrix to start, I obtain for each term: \begin{align} \frac{\partial f}{\partial x^k}\biggr\vert_{F(p)}\frac{\partial^2 F^k}{\partial y^i \partial y^j}(p) &= \frac{\partial f}{\partial x^k}\biggr\vert_{F(p)} \frac{\partial g_{ki}^{-1/2}}{\partial x^m} \frac{\partial F^m}{\partial y^j} = \frac{\partial f}{\partial x^k}\biggr\vert_{F(p)} \frac{\partial g_{ki}^{-1/2}}{\partial x^m} g_{mj}^{-1/2}, \\ -\Gamma^p_{kl}\frac{\partial f}{\partial x^p}\biggr\vert_{F(p)} \frac{\partial F^k}{\partial y^i}(p) \frac{\partial F^l}{\partial y^j}(p) &= - \frac{\partial f}{\partial x^p}\biggr\vert_{F(p)} \Gamma^p_{kl} g_{ki}^{-1/2}g_{lj}^{-1/2} = - \frac{\partial f}{\partial x^k}\biggr\vert_{F(p)} \Gamma^k_{mn} g_{mi}^{-1/2}g_{nj}^{-1/2}, \end{align} but I can't show that: $$ \frac{\partial g_{ki}^{-1/2}}{\partial x^m} g_{mj}^{-1/2} = -\Gamma^k_{mn} g_{mi}^{-1/2}g_{nj}^{-1/2}. $$ I asked about this more specific relation in the question I linked above. Please feel free to correct anything that's wrong.
Thanks a lot for reading!