I'm reading Jech's Set Theory text and one exercise is to show that the Separation Axioms follow from the Replacement Schema. He hints to use $F=\{(x,x):\varphi(x,p)\}$. To show that this is a function, which is a certain sort of class, we need to know that it is a relation such that $(x,y)\in F$ and $(x,z)\in F$ together imply $y=z$. But a relation is defined as a set of $n$-tuples. But this commits us to the idea that the function is a set, not a class. How do we know that $F=\{(x,x):\varphi(x,p)\}$ is a set when we haven't quantified over any set, which besides would just be a use of Separation anyway?
In general I find this class versus set distinction foggy.
The word "relation" in this context just means "class of ordered pairs". So you don't have to prove $F$ is a set to use Replacement. Indeed, this is the entire point of Replacement: it lets you construct sets from functions on set domains, even if you only know those functions are classes (rather than sets).
(More precisely, a "class" is really just a formula $\varphi(x)$ with one free variable in the language of set theory, possibly with parameters. The axiom of replacement is usually stated in terms of formulas $\varphi(x,y)$ in two free variables, such that the relation it defines is a function. That is, for any $x$, there is at most one $y$ such that $\varphi(x,y)$. So to use Replacement directly, you don't literally consider the class consisting of ordered pairs $(x,y)$, but rather just directly talk about the formula $\varphi(x,y)$ you would use to define this class.)