Here is a snippet of the first part of the proof of the Yoneda Lemma. The last line says the square commutes, but I don't get that it commutes. So what has gone wrong in the proof?
Following one direction I get $F(h)\circ F(\alpha) (f)$ while in the direction I get $F(\alpha\circ h)(f)$ where $\alpha:y_2\rightarrow x$
The left-hand morphism is slightly misleading; it's not $h$—rather, it's precomposition by $h$.
To see why the square commutes, let $g \in \mathrm{Hom}(y_2,x)$ (that is $g : y_2 \to x$).
The composite along the left and top of the square is given by $$g \mapsto g \circ h \mapsto F(g \circ h)(f)$$ The composite along the bottom and right of the square is given by $$g \mapsto F(g)(f) \mapsto F(h)(F(g)(f))$$ But $F(h)(F(g)(h)) = (F(h) \circ F(g))(f) = F(g \circ h)(f)$ since $F$ is a contravariant functor.
So the square commutes.