The Gauss-Bonnet expression is independent of the metric

The metric $g_t=(1-t)g_0+t{g}_1$ is smooth and positive definite for every $t\in [0,1]$. Cover a neighbourhood of $D$ by finitely many oriented charts carrying positive $g_0$-orthonormal frames, and on each chart let $A_t$ be the $g_0$-self-adjoint positive bundle map with $g_t(u,v)=g_0(A_{t}u,v)$; the positive square root $A_t^{-1\mathrm{/}2}$ is smooth in the point and in $t$. Applying $A_t^{-1\mathrm{/}2}$ to a $g_0$-orthonormal frame gives a positive $g_t$-orthonormal frame, and because $A_t^{-1\mathrm{/}2}$ acts identically in every chart, the transition functions between overlapping charts are the same $SO(2)$-valued functions for all $t$.

On each boundary component choose a smooth starting point and divide its finitely many $C^2$ arcs further into finitely many chart-contained pieces. Make every genuine corner interior to one chosen frame chart, and put every added chart cut at a smooth point. For each $t$, choose a continuous angle lift ${\theta }_j^t$ of the $g_t$-unit tangent relative to the selected $g_t$-frame on each smooth piece. Let ${\mathrm{Rot}}_t$ be the sum of their angle increments plus the genuine corner jumps ${\alpha }_j(g_t)$ of [F5]. Integrating [F3] piecewise and summing gives ${\int }_{\mathrm{\partial }D}{k}_{g_t}d{s}_{g_t}+{\sum }_j{\alpha }_j(g_t)={\mathrm{Rot}}_t+{\sum }_j{\int }_{\text{piece }j}{\omega }_{t,j}$, where each connection form is taken in that piece's selected frame.

Let ${\omega }_t$ be the connection form of the $g_t$-frame on a chart and set $\beta :={\omega }_1-{\omega }_0$. On an overlap, [F2] gives ${\omega }_t^{\alpha }={\omega }_t^{\beta }+d{\phi }_{\alpha \beta }$ with ${\phi }_{\alpha \beta }$ independent of $t$ by step 1.1, so $\beta$ is chart-independent and defines a global smooth one-form on a neighbourhood of $D$.

At a smooth artificial chart cut, the angle of the same tangent in the next frame differs from its angle in the preceding frame by the negative of their frame-transition angle, modulo $2\pi$. Step 1.1 makes that transition independent of $t$; choose its lift once, so the jump between the two continuous angle lifts is fixed throughout $[0,1]$. At a genuine corner the jump between the one-sided angles in their common frame is the principal exterior angle ${\alpha }_j(g_t)$, with the branch fixed continuously in $t$ because the tangent rays never become antipodal. Thus, on each closed boundary component, ${\mathrm{Rot}}_t$ plus the sum of the fixed artificial-cut jumps is an integral multiple of $2\pi$: after all smooth increments and jumps the unit tangent returns to its starting direction. Both terms are continuous in $t$, and the artificial-cut sum is constant, so this integer multiple is constant. Summing over the boundary components gives ${\mathrm{Rot}}_1={\mathrm{Rot}}_0$.

By [F1], $d{\omega }_t=-K_{g_t}d{A}_{g_t}$ for every $t$, hence $d\beta =d{\omega }_1-d{\omega }_0=-K_{g_1}d{A}_{g_1}+K_{g_0}d{A}_{g_0}$, that is $K_{g_1}d{A}_{g_1}-K_{g_0}d{A}_{g_0}=-d\beta$.

Subtract the two piecewise identities of step 1.2. The connection forms ${\omega }_t$ need not be globally defined, but on every chosen boundary piece their difference ${\omega }_1-{\omega }_0$ is the restriction of the global form $\beta$ from step 2.1. Therefore, by steps 3.1 and 2.2, $G(g_1)-G(g_0)=-{\int }_{D}d\beta +({\mathrm{Rot}}_1-{\mathrm{Rot}}_0)+{\sum }_j{\int }_{\text{piece }j}({\omega }_{1,j}-{\omega }_{0,j})=-{\int }_{D}d\beta +{\int }_{\mathrm{\partial }D}\beta$. Stokes [F4] makes this zero, so $G(g_1)=G(g_0)$.

For the nonorientable closed case, fix a finite curvilinear triangulation of $M$, which exists by [F7] applied to $g_0$; orient each triangular face arbitrarily and give it the induced boundary orientation. Each face is a compact regular disk region with ordinary corners and carries both restricted metrics, so steps 1.1-3.1 apply to it: writing $G^{(f)}(g)={\int }_f{K}_{g}d{A}_g+{\int }_{\mathrm{\partial }f}{k}_{g}d{s}_g+\sum \alpha (g)$ for the face functional, one has $G^{(f)}(g_0)=G^{(f)}(g_1)$ for every face.

Summing over the finitely many faces, the area terms combine to the orientation-free integrals ${\int }_M{K}_{g_0}{\mu }_{g_0}$ and ${\int }_M{K}_{g_1}{\mu }_{g_1}$ by [F9]. Each interior edge has two incident faces on opposite sides; their inward conormals are opposite independently of their arbitrary face orientations, because reversing a face orientation reverses both its boundary tangent and its quarter-turn. Thus the signed curvature integrals cancel, and there are no boundary edges. At each vertex $v$ with $m_v$ incident faces the face-corner angles sum to $2\pi$, so the corner terms contribute ${\sum }_v(m_v\pi -2\pi )$, a number independent of the metric. Hence ${\int }_M{K}_{g_0}{\mu }_{g_0}={\int }_M{K}_{g_1}{\mu }_{g_1}$.