I know how to construct the dual of a statement concerning objects and morphisms of a category, and understand the duality principle associated, but I am having trouble when various categories and functors between them are considered.
For example, this version of the Yoneda lemma (I think it is the dual of the usual one): For every $F:\mathbb{C}\rightarrow \text Sets$ and $C\in Ob(\mathbb{C})$, the following function is a bijection: $$\theta:Hom_{Sets^{\mathbb{C}}}((C,-),F)\rightarrow F(C)$$ $$\theta(\tau)=\tau_C(1_C).$$
What are the steps to construct its dual version? I am really confused about this and appreciate any help you can give me.
This isn't actually dual; this is just the usual Yoneda lemma for functors $\mathbb{D}^{\mathrm{op}}\to\mathbf{Set}$, where $\mathbb{D}=\mathbb{C}^{\mathrm{op}}$, since the functor $(C,-)$ on $\mathbb{C}$ corresponds to the functor $(-,C)$ on $\mathbb{D}^{\mathrm{op}}$.
There is, regardless, a subtlety to the applying the duality principle when functors are involved. Consider the statement "Every faithful functor $F:\mathcal{C}\to\mathcal{D}$ reflects monomorphisms". (Indeed, if $Ff$ is a monomorphism and $fg=fh,$ then $FfFg=FfFh$, $Fg=Fh$, and $g=h$.) What is its dual? There are three places to try to dualize: "faithful", "reflects," and "monomorphisms." Certainly the last will get dualized to "epimorphisms", but what about the other two? It's tempting to switch "faithful" to "full," and it might be tempting to switch "reflects" to "preserves." But in fact when we replace the categories in the statement with their duals, we get a functor $F^{\mathrm{op}}:\mathcal{C}^{\mathrm{op}}\to\mathcal{D}^{\mathrm{op}}$ which is still faithful. And it goes in the same direction as $F,$ so we don't dualize the "reflects" part, either: the dual statement is "every faithful functor reflects epimorphisms." The upshot of all this is that the automorphism $\mathcal{C}\mapsto\mathcal{C}^{\mathrm{op}}$ of $\mathbf{Cat}$ which gives rise to the duality principle is covariant, and in particular doesn't change the functions making up a functor $F$ at all.