The purpose of this answer is to elaborate on the answer by Knzhou. We have the following statement:
Theorem: Suppose that $\delta$ is a general (i.e. not necessarily vertical) quasisymmetry of the action $S$. Then there is an equivalent vertical quasisymmetry $\delta^\ast$, under which the action as also quasi-invariant with the same Noether current.
This statement has two caveats:
- Although the full Noether currents are equivalent, the distribution of the "bare" Noether current and the "improvement term" are not the same in the two cases. In particular if $\delta$ is an exact symmetry, $\delta^\ast$ will be a quasisymmetry in general.
- The corresponding vertical symmetry $\delta^\ast$ is a generalized symmetry in general, even if $\delta$ is an ordinary symmetry. The difference between the two will be explained in the body in the answer.
Suppose that the action is$$S[\phi]=\int_\Omega\mathcal L(x,\phi(x),\partial\phi(x))d^nx,$$ where $\Omega$ is a compact $n$ dimensional domain of integration, the field is $\phi^i(x)$ with $m$ components and $n$ independent variables $x^\mu$. For simplicity, a first order Lagrangian is assumed, but the result is qualitatively valid for the higher order case as well (i.e. the specific formulae are different, but the overall result is the same).
We first consider a variation $\delta$ given by $$ x^{\prime\mu}=x^\mu+\epsilon\delta x^\mu,\quad\phi^{\prime i}(x^\prime)=\phi^i(x)+\epsilon\delta\phi^i(x). $$ We assume this variation is an off-shell quasisymmetry of the action, i.e. $$ \delta S[\phi]=\int_\Omega d_\mu F^\mu\ d^nx$$ for some improvement current $F^\mu$. Let $L=\mathcal Ld^nx$ be the Lagrangian $n$-form and define the total variation $\delta_T\mathcal L$ of the Lagrangian density to be $$ \delta L=\delta_T\mathcal L\ d^nx. $$
As OP has correctly derived we have $$ \delta_T\mathcal L=E_i(L)(\delta\phi^i-\partial_\mu\phi^i\delta x^\mu)+d_\mu\left[\frac{\partial\mathcal L}{\partial\phi^i_\mu}\delta\phi^i-\left(\frac{\partial\mathcal L}{\partial\phi^i_\mu}\phi^i_\nu-\mathcal L\delta^\mu_\nu\right)\delta x^\nu\right] \\ = E_i(L)(\delta\phi^i-\partial_\mu\phi^i\delta x^\mu)+d_\mu\left[\frac{\partial\mathcal L}{\partial\phi^i_\mu}\left(\delta\phi^i-\partial_\nu\phi^i\delta x^\nu\right)+\mathcal L\delta x^\mu\right] \\ = E_i(L)\delta^\ast\phi^i+d_\mu\left[\frac{\partial\mathcal L}{\partial\phi^i_\mu}\delta^\ast\phi^i+\mathcal L\delta x^\mu\right],$$ where $E_i(L)$ is the Euler-Lagrange expression of the Lagrangian, and $$ \delta^\ast\phi^i=\delta\phi^i-\partial_\mu\phi^i\delta x^\mu $$is the vertical part of the variation.
The full Noether current is then $$ J^\mu=\frac{\partial\mathcal L}{\partial\phi^i_\mu}\delta^\ast\phi^i+\mathcal L\delta x^\mu-F^\mu . $$
Now consider making only $\delta^\ast\phi^i$ as a variation with no horizontal part. The variation of the action is $$ \delta^\ast S=\int_\Omega\left(E_i(L)\delta^\ast\phi^i+d_\mu\left(\frac{\partial\mathcal L}{\partial\phi^i_\mu}\delta^\ast\phi^i\right)\right)d^nx. $$ Comparing this with the integral of $\delta_T\mathcal L$, we find $$ \delta^\ast S=\delta S-\int_\Omega d_\mu(\mathcal L\delta x^\mu)d^nx. $$
It follows that if $\delta$ is a quasisymmetry of $S$ with improvement term $F^\mu$, then $\delta^\ast S$ is also a quasisymmetry of $S$ with improvement term $\bar F^\mu=F^\mu-\mathcal L\delta x^\mu$. The full Noether current from this symmetry is then $$ J^\mu=\frac{\partial\mathcal L}{\partial\phi^i_\mu}\delta^\ast\phi^i-\bar F^\mu=\frac{\partial\mathcal L}{\partial\phi^i_\mu}\delta^\ast\phi^i+\mathcal L\delta x^\mu-F^\mu, $$ which agrees with the previous current.
Remarks:
Clearly, if the improvement term vanishes for $\delta$, i.e. $F^\mu=0$, then the corresponding vertical variation still has an improvement term $-\mathcal L\delta x^\mu$. Thus - as stated in the intro - a non-vertical exact symmetry may only be replaced with a vertical quasisymmetry.
Secondly, the variations that appear in Noether's theorem are such that they are not variations about a specific field $\phi^i$, but rather, the variation of any field can be calculated. In other words, $\delta$ is a "vector field" rather than a single "tangent vector" in field space. Ordinarily, these variations have the functional form $$ \delta x^\mu=\xi^\mu(x),\quad\delta\phi^i(x)=\Xi^i(x,\phi(x)). $$ If the variations have this functional dependency, then they generate projectable flows in the $n+m$-dimensional total space of both dependent and independent variables $(x^\mu,y^i)$ (this point would be made much more understandable in a fibre bundle formulation, which I am not doing for the sake of accessibility).
However for vertical variations we may also consider variations of the form $$ \delta\phi^i(x)=\Xi^i(x,\phi(x),\partial\phi(x),...,\partial^r\phi(x)), $$ and this is called a generalized variation (and if a symmetry, then a generalized symmetry) (further remark: One could in principle consider any functional $\delta\phi^i(x)=\Xi^i[\phi](x)$, however in the spirit of locality, one usually considers finite order functionals only).
Generalized vertical variations do not generate flows in the $(x,y)$-space, but they generate flows in the field space, via the equation $$ \frac{\partial\phi^i_\epsilon}{\partial\epsilon}=\Xi^i(x,\phi_\epsilon(x),...,\partial^r\phi_\epsilon(x)). $$
With this in mind, if $\delta$ is an ordinary, non-vertical variation with $$ \delta x^\mu=\xi^\mu(x),\quad \delta\phi^i(x)=\Xi^i(x,\phi(x)), $$ then the corresponding vertical variation has the form $$ \delta^\ast\phi^i(x)=\mathrm Z^i(x,\phi(x),\partial\phi(x))=\Xi^i(x,\phi(x))-\partial_\mu\phi^i(x)\xi^\mu(x), $$ which shows that $\delta^\ast$ is actually a generalized vertical symmetry.
Thus - as stated in the intro - a non-vertical ordinary variation may only be replaced by a vertical generalized variation in general.
