I'm rather new to this subjects so please try to keep it simple.
Let $f:A\rightarrow B$ be a ring homomorphism, and let $N$ be a $B$-module. Then $N$ is an $A$-module as follows: $a\cdot n:=f(a)\cdot n$.
The resulting functor: Rest$_f=$Rest$_{B/A}:$ Mod $B \rightarrow$ Mod $A$, is called restriction functor.
My first question is why is it called that way?
By Wikipedia, it restricts the scalars, but in what way? What is the quotient symbol $B/A$ for?
My second question is how come Rest$_f$ is not Mod $A \rightarrow$ Mod $B$? Where does it take the element $b\cdot n$ when there is no $a\in A$ s.t $b=f(a)$?
Remark (on second question): I get the reason behind how Rest$_f$ works on morphisms Mod $B \rightarrow$ Mod $A$, my problem is only regarding the objects.
First, here $B/A$ is not a quotient. It is just a way of saying "$B$ is over $A$". This is the same kind of notation you might use for a field extension $L/K$ (where obviously the quotient makes no sense).
Then to come to your main question, imagine $A\subset B$ and $f$ is just the inclusion map. Then the functor is quite literally a restriction: you take a $B$-module, and you say "well, if $B$ acts on it, then I can just restrict this action to the subset $A$, and this gives an $A$-module". When $f$ is just injective, this point of view still holds by identifying $A$ and its image. When $f$ is not even injective, then it starts to become more of an image than a factual description, but it still conveys the correct intuition: if we have an action of $B$, then we have an action of $A$ for free.
Your second question seems to show that you don't really understand what this functor does. It takes a $B$-module; so there is no need to define $b\cdot n$ for $b\in B$: it is already defined by assumption. The question is how to define $a\cdot n$ for $a\in A$. (And the solution is to define it as $f(a)\cdot n$, which makes sense since $f(a)\in B$.)