When is a function $f$ transitive?

I'm not used to transitive meaning this, but I can comment on your interpretations.

I believe it's saying that if we have $f:X\to Y$ then $\forall y\in Y,\,f(y)=y$

There are two things wrong with this. Firstly, for generic sets $X,Y$, it makes no sense (it is undefined) to write $f(y)$ for $y\in Y$ when the function $f$ is defined on $X$. It does make sense if, say, $Y\subseteq X$, but not in general. So, we should have $f:X\to X$. The other problem is that you want, for all $y\in Y$, $f(y)=y$. But they only want $f$ to be the identity on its image, so you should have quantified: $\forall y\in Y,\,\exists x\in X,\,y=f(x),\,f(y)=y$ - if you like symbolic descriptions, that is.

Now if we wish to interpret a function as a relation - which is the standard 'implementation' of function - then these posts call functions transitive if the associated relation is transitive. Notice that relations are only said to be transitive when they are also endorelations, i.e. a subset of $X\times X$ for some set $X$. This is another reason why we need $f:X\to X$.

The associated relation $R$ is $(a,b)\in R$ iff. $b=f(a)$. For this to be transitive, $a\sim b$ and $b\sim c$ should mean $a\sim c$. Unwinding definitions, if $b=f(a)$ and $c=f(b)$, then $c=f(a)$. So $f(a)=c=f(b)=f(f(a))$ for all $a\in X$.

Now you should see that the relation induced by $f$ is transitive if and only if the function $f$ satisfies $f\circ f=f$, i.e. it is idempotent. That is the exact same thing as saying "$f$ is the identity on its image."

The comment about $(a,b)\in f$ is abuse of notation, let's rather say, $(a,b)\in R\implies(b,b)\in R$. That's saying $b=f(a)\implies b=f(b)$, which is the same as: $f(f(a))=f(a)$ (for all $a$) hence $f\circ f=f$.