Notations and Variations in the proof for Noether's theorem for fields In the proof for Noether's theorem, as given in D.Gross's notes, there are two kind of variations used.

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*$x'^{\mu}=x^{\mu}+X^{\mu}_\alpha(x)\omega^\alpha$

*$\phi'_i(x')=\phi_i(x)+\Psi_{i\alpha}(x)\omega^{\alpha}$
I understand the first as change in coordinates transformation (translations, Lorentz transformations etc) and the second one as a)internal symmetries b) variation introduced in the field via coordiate transformations (for a non-scalar field the field itself would have a transformation when we perform Lorentz transformations on the coordiates). My question is:

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*How do I understand the various notations used? $\phi_i(x)$ is the field, $\phi_i(x')$ is the field with the transformed coordinates, $\phi_i'(x')$ the transformed field in the transformed coordiantes and $\phi_i'(x)$ is the transfored field in the old coordinates? Can anyone give an explicit example or analogy for this?


*How does makes sense of the variation symbols $\delta\phi_i=\phi_i'(x')-\phi_i(x)$ and $\bar{\delta}\phi_i=\phi_i'(x)-\phi_i(x)$?
Please help! Have been really confused with this.
 A: I have thought of the following analogies. Take the following transformation $x'=x+\alpha$ and the function $f(x)=x^2$. Now I process it as:
$$f(x)=x^2\\f(x')=x'^2\\f'(x')=(1+\alpha)(x'-\alpha)^2\\f'(x)=(1+\alpha)(x-\alpha)^2$$
where $(1+\alpha)$ in the parenthesis is an artificial parameter I have put to mimic what happens to non scalar functions under Lorentz transformations (say). This leads to $\delta f=f'(x')-f(x)=(1+\alpha)(x'-\alpha)^2-x^2=(1+\alpha)x^2-x^2=\alpha x^2$. Also now, $\bar{\delta}f=f'(x)-f(x)=(1+\alpha)(x-\alpha)^2-x^2=\alpha (x-\alpha)^2-(2x-\alpha)\alpha=\alpha x^2-2x\alpha+$ (higher order terms in $\alpha$) . As per the given definitions of the variations, $X^{\mu}_\alpha=1$, $\omega^\alpha=\alpha$, $\Psi_{i\alpha}=x^2$. This relation is known  $$\bar{\delta}\phi_i=\delta \phi_i-\delta x^\mu \partial_\mu \phi_i$$ which when adapted to our case ($\delta x^\mu \partial_\mu \phi_i\sim \alpha \frac{d(x^2)}{dx}=2x\alpha$) looks like $$\bar{\delta}f=\delta f-2x\alpha$$ and is indeed true as per our analysis. Do tell me if my inference and analysis is correct. Thanks!
A: This image might help 
Now onto the 2 points:

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*$\phi(x)$ is the value of old (before transformation) field at the old coordinate point. $\phi'(x')$ is the value of new (after transformation) field at the new coordinate point. As an example, transformation is a simple displacement, $x\rightarrow x'=x+2$. So $\phi(2)$ stands for value of $\phi$ at the coordinate value $2$ in the old coordinate system. While $\phi'(2')$ stands for the value of the transformed field at the coordinate value $2$ in new system which is $0$ in old coordinate system. $\phi'(2)$ stands for the value of transformed field at coordinate value $2$ in old system which is $4'$ in the new system. A simple mnemonic I use old(old), old(new), new(old), new(new) to remember these transformed value.

*$\delta(\phi)$ means change in value of field when compared at same point but from different coordinate system (From the above image it is $0$ for a scalar field). While $\bar{\delta}\phi$ is the difference of value of new field at one place and value of old field at different place. Though the tag holder for the place seems same but remember value feed into new field has different location. Explicit example $$\phi'(2)-\phi(2)$$
$$=\phi'(4')-\phi(2)$$ $\bar{\delta}\phi$ is used for calculating generators

The issue with your answer is since you've used scalar field and a simple translation transformation. The second term in the field transformation becomes trivially $0$. Try looking at the image once again. $f'(x')=f(x)$ so $\delta f=0$ and $$\bar\delta f=f'(x)-f(x)$$
$$=f'(x'-\alpha)-f(x)$$
$$=-\alpha\frac{\partial f'}{\partial x'}$$
$$=-\alpha \frac{\partial f}{\partial x}$$
Where the last line is not trivial but you can deduce it from last second line since we are only interested in keeping terms of order $1$ in $\alpha$.
Try looking in yellow book of CFT (ch$-1$) for further discussion. Also if you want to see the second term explicit you have to go for tensorial or spinor field cause in case of scalar for a Poincare transform $\phi'(x')=\phi(x)$
