How to compute the variation of functionals During my physics classes I encountered various definitions on how to calculate the variation of a functional/function for certain "boundary conditions" and the thing is, I don't really understand if they are all equal or if I have to use some in certain situations... 
Lets assume I have a functional of the form 
$$S[f] = \int_{x_i}^{x_f} d^4x\, \mathcal{L}(f(x), \partial_\mu f(x),x).$$
The first definition I ever saw was
$$\delta S := \frac{d}{d \varepsilon} S[f+ \varepsilon \delta f]\Big|_{\varepsilon =0}\quad \text{where}\quad \delta f(x_i)=\delta f(x_f)=0.\label{1} \tag{1}$$
Sometime later I found 
$$\delta \mathcal{L}= \delta_o\mathcal{L} +\delta x^\mu \partial_\mu \mathcal{L}\label{2}\tag{2},$$
where
$$\delta x^\mu = (x^\mu)'-x^\mu,\quad \delta_o f= f'(x)-f(x),\quad \delta f = f'(x')-f(x),$$
for $x\mapsto x'$ some kind of transformation. I also saw 
$$\delta\mathcal{L} = \frac{\delta \mathcal{L}}{\delta f}\delta f + \frac{\delta \mathcal{L}}{\delta(\partial_\mu f)}\partial_\mu \delta f \quad\text{where} \quad [\delta, \partial_\mu]=0\label{3}\tag{3}$$
was stated without any explanation.

The only thing that the above "$\delta$" all had in common was, that no at single of the professors really explaind what I was supposed to do with them.
TL;DR: How is this $\delta$ defined, is $\delta \mathcal{L}$ different from $\delta S$ and $\frac{\delta \mathcal{L}}{\delta f}$? And if someone tells me to calculate $\delta \mathcal{L}$, how am I supposed to do this? 
Preferably, I'd like to see an example for a global symmetry and a local symmetry transformation and how to calculate $\delta\mathcal{L}$ (it seems that different rules apply here). 
 A: Premises 
For the reasons briefly exposed in this Q&A,  you should not rely too much on the Wikipedia entry on functional derivatives: in the sequel, I also assume (just for reference, since this choice does not significantly impact on the mathematical developments involved) that the integral functional $S(f)$ is defined as follows
$$
S[f] = \int\limits_{x_i}^{x_f}\!\! \mathcal{L}\big(x,f(x), \partial_\mu f(x)\big) \mathrm{d}^4x\,\label{A}\tag{A}
$$
Said that, I'll try to give ananswer to your questions, following mainly the two original source [1]. 
Questions and answers

TL;DR: How is this $\delta$ defined, is $\delta \mathcal{L}$ different from $\delta S$ and $\frac{\delta \mathcal{L}}{\delta f}$? And if someone tells me to calculate $\delta \mathcal{L}$, how am I supposed to do this?

Strictly speaking, $\delta$ represents the increment of a given quantity defined on a function space, say $X$ (it could be $X=C^k([a,b])$, for example). 


*

*If $f\in X$ is intended as an independent variable, then the variation of $f$ is defined as ([1] §I.II.1.26 p. 22)
$$
\delta f(x)=\varepsilon \varphi(x) \quad \varepsilon>0\label{B}\tag{B}
$$
where $\varphi\in X$ as well. 

*In a similar, but not identical, way if $S: X\to \Bbb R$ is a functional (or an operator between function spaces $X\to Y$ like $\mathcal{L}$), its variation is defined as ([1] §I.II.1.28 p. 24)
$$
\begin{split}
\delta S(\delta f)=\delta S(\varepsilon\varphi) & = S(f+\delta f) - S(f)\\
& = S(f+\varepsilon\varphi) - S(f)
\end{split}\quad\left(
\begin{split}
\delta \mathcal{L}(\delta f)=\delta \mathcal{L}(\varepsilon\varphi) & = \mathcal{L}(f+\delta f) - \mathcal{L}(f)\\
& = \mathcal{L}(f+\varepsilon\varphi) - \mathcal{L}(f)
\end{split}\right)
$$
Clearly, if we assume \eqref{A}, we have $\delta S(\varphi)\neq\delta \mathcal{L}(\varphi)$

*The functional derivative of a functional (or an operator) is defined as
$$
\begin{split}
\frac{\delta S}{\delta f}= \frac{\delta S}{\delta f}(\varphi) &:= \frac{\mathrm d}{\mathrm d \varepsilon} S[f+ \varepsilon \varphi]\Big|_{\varepsilon =0}\\
&=\lim_{x\to 0}\frac{S(f+\delta f) - S(f)}{\varepsilon}\\
&=\lim_{x\to 0}\frac{S(f+\varepsilon\varphi ) - S(f)}{\varepsilon}
\end{split}\label{1'}\tag{1'}
$$
In your equation \eqref{1}, $\varphi(x)\equiv\delta f(x)$ by (though customary) abuse of notation: also, the condition $\delta f(x_i)=\delta f(x_f)$ is only required when your aim is to deduce the Euler-Lagrange equations for a functional of integral type like $S(f)$ \eqref{A} is.


From the simple examples given, but more properly from the definition \eqref{1'}, it is clear that the functional derivative is the extension to function spaces of the directional derivative (and thus also of the partial derivative) concept: as we can see, if we assume $X=\Bbb R^n$, then we have that $\varphi(x)=(0,\ldots,0,x^\mu,0\ldots,0)$ we have
$$
\delta f(x)=\varepsilon(0,\ldots,0,x^\mu,0\ldots,0):=\delta x^\mu=\partial x^\mu\label{C}\tag{C}
$$
I think this is the problem behind the confusion in the example shown: by abuse of notation, the same $\delta$ symbol is adopted for both functionals and variations defined on finite and infinite dimensional spaces, mixing up the two (however closely related) concepts. Thus


*

*$\delta x^\mu= (x^\mu)'-x^\mu$ is the variation of an independent variable in a finite dimensional space, in the spirit of \eqref{C}: the $\varepsilon$ parameter is implicitly considered in the transformation $(\:)^\prime:\Bbb R^n \to \Bbb R^n$ (if we are working on Euclidean spaces).

*$\delta_o f=f^\prime(x) -f(x)$ is the variation of an independent variable in a function space, in the spirit of \eqref{B}, and again the $\varepsilon$ parameter is implicitly considered in the transformation $(\:)^\prime:\Bbb R^n \to \Bbb R^n$.

*Finally, $\delta f=f^\prime(x^\prime)-f(x)$ is a mix of the two variation concepts \eqref{B} and \eqref{C} in the sense that in this case the function varies both as an element of a function space and bot as a function of its independent variable $x$. 


These observation give meaning to the commutator relation $[\delta,\partial_\mu]=0$, and perhaps, by using the formulas above it seems possible to deduce \eqref{3} from \eqref{2}.
Unfortunately I cannot offer you an example calculation involving the formalism of symmetries as used by physicists, since I am not accustomed to it.
References
[1] Volterra, Vito, Theory of functionals and of integral and integro-differential equations. Dover edition with a preface by Griffith C. Evans, a biography of Vito Volterra and a bibliography of his published works by Sir Edmund Whittaker. Unabridged republ. of the first English transl, New York: Dover Publications, Inc. pp. 39+XVI+226 (1959), MR0100765, ZBL0086.10402.
A: Well I can see that your first and third definitions are compatible.But the second one I don't understand.
Start from your first definition:
$$
\delta S = \frac{d}{d\varepsilon}(f+\varepsilon \delta f) = \lim_{\varepsilon\rightarrow0}\frac{1}{\varepsilon}\int_{x_i}^{x_f}\left[\mathcal{L}((f+\varepsilon \delta f),\partial_{\mu}(f+\varepsilon \delta f),x)-\mathcal{L}(f,\partial_{\mu}f,x)\right]d^4x = \int_{x_i}^{x_f}(\frac{\partial \mathcal{L}}{\partial f}\delta f+\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}f)})\delta (\partial_{\mu}f))d^4x 
$$
Because the operator $\delta$ performs on $f$ but the operator $\partial_{\mu}$ performs on $x$, $[\delta,\partial_{\mu}]=0$.
Then from your third definition: $\delta S= \int_{x_i}^{x_f}\delta \mathcal{L}d^4x$
