Does $y = f(x) = ax+b$ actually have two mappings inside it? I’m just a high school student, so I may be somewhat logically flawed in understanding this.
According to wikipedia, the definition of function requires an input $x$ with its domain $X$ and an output $y$ with its domain $Y$, and the function $f$ maps $x$ to $y$.
But how about $f(x)$? I often see syntaxes such as $f(1) = 0$  in my text book. Doesn’t that mean it is $f(x)$ being first assigned a value and then transfer the value into $y$? So, there must be two transitions/mappings between the input $x$ and the output $y$ right?
My conceptual model of function is like this: A definition of function requires an input $x$ with its domain $X$, a forwarder $f(x)$ with its domain $F$ and an output $y$ with its domain $Y$. The function $f$ first maps $x$ to $f(x)$ then maps $f(x)$ to $y$.
These two definitions are not quite the same.
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
&quot;the number that $f$ maps $1$ to.&quot;
So if you ask, &quot;What is $f(1)$,&quot; 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 &quot;transitioning&quot; 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, &quot;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 ... .&quot; 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.
I see what you are saying, but there are ways to think about this without using more than one mapping, as the other answers have pointed out.
But here is what I think you are getting at (which I agree with):
$x \mapsto f(x) \mapsto ax+b$. where we have a general function of $x$ and we send it to some expression in terms of the variable $x$.
Then $x$ maps to $f(x)$ as normal by using the domain and codomain.