Why do we pretend + and x not to be addition and multiplication while eventually they are considered so while explaining groups, rings and fields? Aren't commutative, associative properties result of addition and multiplication? Then why most of the definitions of Groups and Fields contain + and x but don't accept that they are addition and multiplication?
 A: 
[…] but don't accept that they are addition and multiplication?

This is a wrong deduction:
When defining algebraic structures, one often refers to the used binary functions as some kind of “addition” or “multiplication”, to build up an analogy to the actual, “standard” addition and multiplication. These terms are not part of the definition, although many books use these to provide further explanation and give the reader a better expression about the meaning of these functions.
For instance, one can define a group to be a duple $(G, \cdot)$ where:


*

*$G$ is some set

*$\cdot:G\times G\rightarrow G$ a binary operation

*$\cdot$ is associative

*$\exists e\in G\forall g\in G: g\cdot e = e\cdot g = g$ (existence of a neutral element)

*$\forall g\in G \exists g^{-1}\in G: g\cdot g^{-1}=e$ (existence of an inverse element)
To get where these terms are mostly used, let's define a field to be a triple $(F, +, \cdot)$ where:


*

*$F$ is a set

*$(F, +)$ is a commutative group

*$(F\setminus\{0\}, \cdot)$ is a commutative group with $0$ being the neutral element of $(F, +)$

*$\forall a,b,c\in F: a\cdot(b+c)=(a\cdot b)+(a\cdot c)$ (distributivity)
But note the following things:


*

*Just because I chose the symbols $\cdot$ and $+$, this does not mean that these are the standard addition and multiplication functions! We just defined them to be functions on an arbitrary set $F$ that satisfy the given conditions.

*Standard addition and multiplication over $\mathbb{R}$ indeed satisfy this definition, which means that $(\mathbb{R}, +, \cdot)$ is an example of a field — but in this case, the used symbols refer to these standard functions. One could call the multiple usage of these operators a “definition overload”. This is not a wrong behaviour; in order to use standard addition and multiplication inside a definition, one would rather say this explicitly — for instance, “Let $+$ denote addition over the reals” — but only if it is dubious. In this case, because the set can be anything, these functions could not possibly refer to addition and multiplication, because they are only defined over the reals, and not any arbitrary set.


But: Why would one do this?
There is a rather basic answer to this: To hold things general. The advantage of this broad definition, which is actually just a property, is that if you show that certain things hold for this structure, it automatically applies for every instanc of this structure — one would only have to show that a certain element is an instance of that structure.
The simplest example for this is the uniqueness of the neutral element of a group — because one already proved this for groups in general, this automatically holds for addition and multiplication as well but also for every other binary operation on a set satisfying the properties of a group!
This principle of keeping things broad holds for even larger things, for example vector spaces. But i must admit that my experience is not large enough to give a really impressive example. I would appreciate if someone could add such in a comment ;)
I hope I cleared up a few things here.
A: As an exemple of how '' addition'' and ''multiplication'' can be different from the usual operations in $\mathbb{R}$, take the ring $(M_2(\mathbb{R},\oplus, \odot)$ of $2\times 2$ matrices with real entries.
Here the $\oplus$ is defined as
$$
A\oplus B=
\left(
 \begin{array}{ccccc}
a_1&a_2 \\
a_3&a_4  
\end{array}
 \right) \oplus
\left(
 \begin{array}{ccccc}
b_1&b_2 \\
b_3&b_4  
\end{array}
 \right) =
 \left(
 \begin{array}{ccccc}
a_1+b_1&a_2+b_2 \\
a_3+b_3&a_4+b_4  
\end{array}
 \right) 
$$
and we can easly see that this operation is associative, commutative, has a neutral element (the matrix with all $0$ entries, that we call $0$) and that every matrix $A$ has an ''inverse $X$'' (that we call ''opposite'') such that $A \oplus X=0$ ( and we call this opposite $-A$), i.e. $(M_2(\mathbb{R}),\oplus)$ is a commutative group.
Is all that sufficient to call $\oplus$ a ''sum'' in analogy whit the sum in $\mathbb{R}$? NO, because also the product in $\mathbb{R}$ has the same formal properties.
But we also define the operation $\odot$ as:
$$
A\odot B=
\left(
 \begin{array}{ccccc}
a_1&a_2 \\
a_3&a_4  
\end{array}
 \right) \odot
\left(
 \begin{array}{ccccc}
b_1&b_2 \\
b_3&b_4  
\end{array}
 \right) =
 \left(
 \begin{array}{ccccc}
a_1b_1+a_2b_3&a_1b_2+a_2b_4 \\
a_3b_1+a_4b_3&a_3b_2+a_4b_4  
\end{array}
 \right) 
$$
and we see that such opertion is associative, but not commutative, there is a neutral element, the matrix
$$
I=
\left(
 \begin{array}{ccccc}
1&0 \\
0&1  
\end{array}
 \right)
$$
but there are matrices $M \ne 0$ that don't have an ''inverse'', as:
$$
M=
\left(
 \begin{array}{ccccc}
0&m \\
0&0  
\end{array}
 \right)
$$
So $(M_2(\mathbb{R}),\odot)$ is not a group but only a monoid.
So the first operation $\oplus$ has the same formal properties of $+$ and $\cdot$ in $\mathbb{R}$, but the second operation $\odot$ has not these properties. Why ve call the first a ''sum'' and the second a ''product''? Because there is another important property: $\odot$ is distributive over $\oplus$, i.e.:
$$
A\odot(B\oplus C)=a\odot B \oplus A\odot C
$$
And here $\oplus$ and $\odot$ have the same behavior as $+$ and $\cdot$.
So $\oplus$ is very similar to the $+$ operation on $\mathbb{R}$ and we can call it a ''sum''. Finally we chose to  call $\odot$ a ''product'' ( or ''multiplication'')  only to save words, but we we must be careful that this ''product'' is not commutative and not invertible.
