In Section 1.6 of Condensed Matter Field Theory by Altland and Simons, they prove Noether's theorem. In order to do so, they consider an infinitesimal transformation of the coordinates and the field:
$$ \begin{align} x_\mu \to x'_\mu =& x_\mu + \left.\frac{\partial x_\mu}{\partial \omega_a}\right|_{\omega = 0} \omega_a(x), \cr \phi^i(x) \to \phi'^i(x') =& \phi^i(x) + \omega_a(x) F^i_a[\phi]\end{align}\tag{1.42} $$
to linear order in a set of parameter functions $\{\omega_a\}$ characterizing the transformation, where the functionals $\{F^i_a\}$ can depend on the coordinate $x$ and define the incremental change $\phi'(x') - \phi(x)$.
They then compute the change $\Delta S$ in the action to linear order in $\omega_a$ as
$$ \Delta S = \int d^m x'\, \mathcal{L}(\phi'^i(x'), \partial_{x'_\mu}\phi'^i(x')) - \int d^m x \,\mathcal{L}(\phi^i(x), \partial_{x_\mu} \phi^i(x)) $$
$$ = \int d^m x\, (1 + \partial_\mu (\omega_a \partial_{\omega_a} x_\mu)) \mathcal{L}(\phi^i + F^i_a \omega_a, (\partial_{\mu \nu} - \partial_\mu (\omega_a \partial_{\omega_a} x_\nu)) \partial_\nu (\phi^i + F^i_a \omega_a)) - \int d^m x \, \mathcal{L}(\phi^i(x), \partial_\mu \phi^i(x)). $$
I calculated this myself and got the same answer as them. Up to this point, they do not use the fact that the transformation is a symmetry transformation. Next, they say if the transformation is a symmetry, then for constant parameters $\omega_a$, the action difference $\Delta S$ vanishes. Then, the remaining terms in $\Delta S$ are proportional to $\partial^\mu \omega_a$ (which vanishes when $\omega_a$ is constant) and are given by
$$ \Delta S = -\int d^m x \, j_\mu^a(x) \partial^\mu \omega_a,\tag{p.33} $$
where the Noether current is given by
$$ j^a_\mu = \left.\left(\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi^i)}\partial_\nu \phi^i - \mathcal{L} \delta_{\mu \nu}\right)\frac{\partial x_\nu}{\partial \omega_a}\right\vert_{\omega = 0} - \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi^i)} F^i_a.\tag{1.43} $$
They then claim that since the linear variation of the action in any parameter must vanish on-shell (because the equations of motions hold classically), it follows that $\partial^\mu j_\mu^a = 0$.
I tried working this out by myself, and it seems to me like this only holds in general if the symmetry is a symmetry of the Lagrangian and does not hold in general if the symmetry is only a symmetry of the action, because in this case, we would have to add additional terms to the Noether current due to boundary terms. Is this correct, or am I missing something that allows their proof to work for symmetries of the action as well?
