lagrangian formalism - Noether's theorem with infinite parameters

I'm trying to understand something regarding Noether's theorem - and with the given situation, my question isn't that much of a question, I'm rather just seeking confirmation whether I'm thinking right or not.

The situation:

Let $\mathcal L$ be a Lagrangian density, depending on some field $\phi$, and its first derivative. Noether's theorem (naively) says that if $\phi(x)\mapsto\phi(x)+\epsilon\delta\phi(x)$ is a specific infinitesimal deformation of the field (a more precise thing would be to say that this is a smooth 1-parameter family of finite deformations - and we're interested in behaviours under $d/d\epsilon|_{\epsilon=0}$), such that $\mathcal L$ changes by a divergence ($\delta\mathcal L=\partial_\mu K^\mu$), then the current $$j^\mu=\frac{\partial\mathcal L}{\partial(\partial_\mu\phi)}\delta\phi-K^\mu$$ is conserved on-shell.

It is known that this can be recast in a different form, by making $\epsilon$ be a function instead of a parameter. Then the action won't be invariant in general, but the deformation of the action gives the same current $j^\mu$, and its conservation can be shown.

The problem is that this doesn't make sense, imo. To show this, consider the following, let $\epsilon(x)$ be the "infinitesimal" functional parameter, the variation is $$\phi(x)\mapsto\phi(x)+\epsilon(x)\delta\phi(x).$$ Let us define $\epsilon'(x)$ and $\epsilon$ as $\epsilon(x)=\epsilon\epsilon'(x)$, where here only $\epsilon$ is "infinitesimal". Now the variation has the form $$\phi(x)\mapsto\phi(x)+\epsilon\epsilon'(x)\delta\phi(x).$$ Now we redefine $\delta\phi(x)$ to $\delta\phi'(x)=\epsilon(x)\delta\phi(x)$, then the variation has the form $$\phi(x)\mapsto\phi(x)+\epsilon\delta\phi'(x).$$

This is literally the same form we had before we assumed $\epsilon$ is a function.

So this begs the question - what do we mean on an "infinite-parameter" variation? The problem is clearly caused by the fact, that if $\delta\phi(x)$ is specific, but reasonably arbitrary, then this still contains as many "free parameters" as the different possible values for $x$. Essentially, $\delta\phi(x)$ already contains infinite parameters.

The resolution:

Looking at specific examples, such as an $U(1)$ transformation of the free, massive, complex Klein-Gordon field, the finite transformation is $$\phi(x)\mapsto e^{i\epsilon}\phi(x).$$ Infinitesimally, this is $$\phi(x)\mapsto \phi(x)+i\epsilon\phi(x),$$ so $$\delta\phi(x)=i\phi(x).$$

Here we see, that $\delta\phi(x)$ depends on $x$ only through the unperturbed field $\phi(x)$ itself, so here the variation is truly 1-parameter.

If we do this for another archetypical example - spacetime translations - we get the same results.

The question:

Am I right in saying that the usual form of the Noether's theorem should be stated that we consider variations of the form $$\phi(x)\mapsto\phi(x)+\epsilon\delta\phi[\phi(x),\partial\phi(x)],$$ where $\delta\phi$ is a specific function of the field $\phi$, and possibly, its derivatives, but not the coordinates $x$?

Because only then does it makes any sense to me to discuss whether the variation has finite or infinite amount of parameters.