Does $y = f(x) = ax+b$ actually have two mappings inside it?
Equality is not just a mapping. It means two things are one and the same.
We wrote $f(x) = y$ not to indicate that we are mapping $f(x)$ to $y$; it means that $f(x)$ is $y$.
Consider a function $f$ that maps any value $x$ in the domain $\mathbb Z$ (the integers) to a value $y$ in the codomain $\mathbb Z$. (Codomain rather than domain is the correct word here.) In particular, $f$ is the function that maps each integer to the previous integer, so $f$ maps $3$ to $2$, $f$ maps $2$ to $1,$ and $f$ maps $1$ to $0.$
Then the notation $f(1)$ is simply a way of saying "the number that $f$ maps $1$ to." So if you ask, "What is $f(1)$," the answer is $0$, because $f$ maps $1$ to $0$.
Again: $f$ maps $1$ to $0$, therefore the number $f$ maps $1$ to is $0$; that is, $f(1)$ is $0$. We express this in the equation, $f(1) = 0$.
It is certainly possible to set up a situation in which a member of one set is mapped to something which is then mapped to something else. But that is not how a single function is defined. A function defines exactly one transition from one set to another, and the notation $f(x)$ is one way to express the final result of that transition. If $f$ maps from a domain $X$ to a codomain $Y$ then $f(x)$ is already a member of $Y$ and needs no further "transitioning" to get to $Y$.
Notice that in the previous paragraph I never mentioned any variable named $y$. Indeed I too take issue with the part of the Wikipedia article that says, "this relation is denoted by $y = f (x)$ ... where the element $x$ is the argument or input of the function, and $y$ is the value of the function ... ." The symbol $y$ is completely superfluous here; if we use $x$ to represent the input of the function, the usual denotation for the value of the function is $f(x)$, and a denotation that actually shows the relation of an element $x$ in the domain to an element in the codomain is $x \mapsto f(x).$ Here $f(x)$ is the element of the codomain and $\mapsto$ is the relation.