What is the meaning of the field deformation (delta phi) for continuous symmetries?

I'm taught in class that for a symmetry $$\phi \rightarrow \phi + \delta\phi$$ (and leaving the spacetime coordinates alone), the Noether current is $$J^\mu = \frac{\partial L}{\partial (\partial_\mu\phi_n)} \delta\phi$$ but the expression $$\delta\phi$$ is very unclear to me exactly what it means. Sometimes, $$J^\mu$$ is written as $$J^\mu = \frac{\partial L}{\partial (\partial_\mu\phi_n)} \frac{\delta\phi}{\delta\alpha}$$ if the symmetry depends on a parameter $$\alpha$$. This makes slightly more sense, but it still makes me uncomfortable because of the following reason. For the U(1) symmetry $$\phi \rightarrow e^{i\alpha}\phi$$, if I blindly take a derivative with respect to $$\alpha$$, then I get $$i e^{i\alpha}\phi$$. However my textbook clearly states that the expression $$\frac{\delta\phi}{\delta\alpha} = i\phi$$ (without the extra phase).

So my question is: is it at least correct (though less concise) to say that a symmetry $$\phi'(\phi,\alpha)$$ is a function of the field and the parameter $$\alpha$$ such that $$\phi'(\alpha=0) = \phi$$, and that $$J^\mu = \frac{\partial L}{\partial (\partial_\mu\phi_n)} \frac{\partial \phi'}{\partial \alpha}\Big|_{\alpha=0}$$ is the Noether current? My concern with this is, what about symmetries that are not parameterized by some parameter $$\alpha$$? What would be a more clear definition of $$\delta\phi$$ in that case?

The transformation $$\phi\rightarrow\phi+\delta \phi$$ is assumed to be small (infinitessimal). You can write $$\delta\phi$$ as \begin{align}\delta\phi&=\phi'-\phi\\ &=e^{i\delta \alpha}\phi-\phi\\ &\approx(1+i\,\delta\alpha)\phi-\phi\\ &=i\,\delta\alpha\,\phi \end{align} So this means $$\frac{\delta\phi}{\delta\alpha}=i\phi$$. Why does this coincide with $$\left.\frac{\partial\phi'}{\partial\alpha}\right|_{\alpha=0}$$? This is exactly because $$\left.\frac{\partial\phi'}{\partial\alpha}\right|_{\alpha=0}$$ gives the first order in a Taylor expansion of $$\phi'$$ in $$\alpha$$ around zero: $$\phi'(\alpha)=\phi'(0)+\alpha \left.\frac{\partial\phi'}{\partial\alpha}\right|_{\alpha=0}+\mathcal O(\alpha^2)$$ Note that $$\phi'(0)=\phi$$. Where does this Taylor expansion appear in the derivation of $$\delta\phi$$?
So to answer your question, yes it would be correct to write $$\phi'$$ like you did. What about symmetries that can't be parametrised by some parameter $$\alpha$$? We are talking about continuous symmetries and I'm not 100% sure but I think all continuous symmetries can be parametrised that way.