So, the whole topic is new to me, sorry if I may be unclear with what I want to solve.
Let $\langle F(U),\circ \rangle$ be an algebraic structure with $F = \left\{f \mid f: U\mapsto U\right\}$. Is the composition associative?
I think yes, the proof of associative composition is quite common but I don't know if it is true in this case with $F$.
Yes, composition of functions is always associative. You can check this directly from the definition: to check that $(f\circ g)\circ h=f\circ(g\circ h)$, you need to check that these two functions agree on any input. So fix $x\in U$ and compute using the definition of $\circ$: $$((f\circ g)\circ h)(x)=(f\circ g)(h(x))=f(g(h(x)))$$ $$(f\circ(g\circ h))(x)=f((g\circ h)(x))=f(g(h(x)))$$