Edit: Reviving this thread because I still could not prove or find a proof of this. A sketch of a proof attempt can be found in the previous edit of this post.
I am trying to show the following extension of the Du Bois Raymond lemma:
Let $M$ be a smooth Riemannian Manifold and $\omega: [0,1] \rightarrow M$ be a $W^{1,2}$ curve on M. Consider a tangential $L^2$ vector field along $\omega$ denoted by $v \in L^2(\omega^*TM)$.
If $$\int_{0}^{1} \langle v, \frac{Du}{\partial t} \rangle \text{ dt} = 0 \ \ \ \ \ \text{for all } u \in W ^{1,2}(\omega^*TM) \text{ with } u(0)=u(1)=0$$
Then
$$v \in W^{1,2}(\omega^*TM)\ \ \text{ with }\ \ \frac{Dv}{\partial t} = 0 \ \ a.e.$$
where $\frac{Du}{\partial t}$ denotes the covariant derivative of $u$ along the curve $\omega$.
For my purposes it would be sufficient to show this for the simplified case where $M = S^2$ and the covariant derivative becomes the projection onto the respective tangent space, i.e., $\frac{Du}{\partial t} = u' - \langle u', \omega \rangle \omega$.
Kind regards.
Expanding on Amitai Yuval's comment...
Fact. A function $g\in L^2((0,1),\mathbb R)$ that satisfies $\int gf=0$ for all compactly supported smooth $f:(0,1)\to\mathbb R,$ must satisfy $g\equiv 0$ a.e.
See for example Haim Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, Corollary 4.24.
More generally, for a finite dimensional inner product space $V,$ a function $g\in L^2((0,1),V)$ satisfying $\int \langle g,f\rangle=0$ for all $f\in C^\infty_c((0,1),V)$ must be zero a.e. - just pick a basis and apply the Fact to each component. So a function $g\in L^2((0,1),T_{\omega(0)}M)$ that satisfies $\int \langle g,f\rangle=0$ for all compactly supported $f\in C^\infty_c((0,1),T_{\omega(0)}M)$ must satisfy $g\equiv 0$ a.e.
Parallel transport gives an isometry of the fiber at $t$ to the fibre at $0,$ in symbols $T_{\omega(t)}M\cong T_{\omega(0)}M,$ and these give a smooth isometric trivialization $\omega^*TM\cong [0,1]\times T_{\omega(0)}M.$ So any function $g\in L^2(\omega^*TM)$ that satisfies $\int \langle g,f\rangle=0,$ for all $g\in C^\infty(\omega^*TM)$ supported on a compact subset of $(0,1),$ must satisfy $g\equiv 0$ a.e.
Bringing this back to the original problem, we are given $v\in W^{1,2}(\omega^*TM)$ such that
$$\int_0^1 \langle v,\frac{Du}{\partial t}\rangle =0$$
for all $u\in C^\infty(\omega^*TM)$ supported on a compact subset of $(0,1).$ By one definition of the weak derivative $Dv/\partial t,$
$$\int_0^1 \langle \frac{Dv}{\partial t},u\rangle=-\int_0^1 \langle v,\frac{Du}{\partial t}\rangle\tag{*}$$ for such $u,$ but we've just said this is zero. So $Dv/\partial t=0$ a.e.
I'm assuming (*), but you might be working with a different definition where this isn't immediate. (*) is the special case of the definition $\int_0^1 \langle \nabla v,u\rangle=-\int_0^1 \langle v,\nabla u\rangle$ for weak derivatives on vector bundles with connection $\nabla$ (the covariant derivative along a curve is the covariant derivative defined by the pullback connection, along the standard vector field $d/dt$).