Given $f,g$ as morphisms of objects in $C$, we denote them as $f',g'$ for $C^{op}$, and $$f\circ_C g = g'\circ_{C^{op}} f'.$$ For a contravariant functor $F$ from $C$ to $D$, we have $f:x\rightarrow y, g: y\rightarrow z$ $$F(g\circ_C f) = F(f) \circ_D F(g)$$ Why is this the same as a functor $F$ from $C^{op}$ to $D$? So given $$F(g' \circ_{C^{op}} f') = F(g')\circ_D F(f')$$ why is $$F(f) \circ_D F(g)= F(g')\circ_D F(f')?$$
Edit: If what I wrote doesn't make sense, can you point me to a reference? I see people say "the opposite category is used to define contravariant in a uniform way", but there is not any detail.
Using your notation, note that $$F(g' \circ_{C^{\mathrm{op}}} f') = F(f \circ_C g)$$ and not $F(g \circ_C f)$, and so you're trying to compare the wrong $D$-morphisms on the 8th line of your question (i.e. the equation after 'why is').
Treating $F$ as a covariant functor $C^{\mathrm{op}} \to D$, we have $$F(g \circ_C f) = F(f' \circ_{C^{\mathrm{op}}} g') = F(f') \circ_D F(g')$$ which is the same result as when we considered $F$ as a contravariant function $C \to D$, since $f=f'$ and $g=g'$.