Why didn't dualization reverse all arrows in this case?

The way the duality principle is phrased is meant to give some initial intuition, but is not suitable (and, considering the qualification as 'informal', not intended to be used) for formal application. One reason you've already mentioned: There are categorical constructions where the 'dual' is not obtained by reversing 'all' arrows. Secondly, while the introductory paragraph can be read to suggest this, dual constructions or proofs need not be carried out 'again', but they are in fact formal consequences of the original construction.

An attempt for a formal statement of the duality principle would be:

For every generic statement $\forall{\mathscr A}: {\mathsf P}({\mathscr A})$ about categories, formulated in some fixed context (perhaps involving other categories) categories, there is a 'dual' generic statement $\forall{\mathscr A}: {\mathsf P}({\mathscr A}^{\text{op}})$, which is equivalent to the original one.

Note that the context of the statement stays the same - dualization applies to the single category the statement is parametric over. In your example of the Yoneda Lemma, the parameter is ${\mathscr A}$, and the statement is that a concrete canonical assignment describes a functor ${\mathscr A}\to [{\mathscr A}^{\text{op}},\textsf{Set}]$. The dual statement hence says that for any ${\mathscr A}$ there is a concrete canonical assignment describing a functor ${\mathscr A}^{\text{op}}\to [{\mathscr A}^\text{op op}(\equiv {\mathscr A}),\textsf{Set}]$.

Duality doesn't work the way it seems you think it does. The point is that if $\mathcal{A}$ is a category, so is $\mathcal{A}^{op}$, and if something is true of every category, then it is also true of the dual of every category. The reason we don't reverse everything is because the theorem above is a theorem about an arbitrary category $\mathcal{A}$, not about an arbitrary diagram $\mathcal{A}\to \mathrm{whatever}$.

Here, the statement is that for any category $\mathcal{A}$, there exists a functor $H^*:\mathcal{A}^{op}\to [\mathcal{A},\mathbf{Set}]$. Since it's true for every category, it's also true for $\mathcal{A}^{op}$; that is, there's a functor $\mathcal{A}=(\mathcal{A}^{op})^{op}\to [\mathcal{A}^{op},\mathbf{Set}]$ taking $a\in \mathcal{A}$ to $$\mathcal{A}^{op}(a,-):\mathcal{A}^{op}\to\mathbf{Set}$$ which, by the definition of $\mathcal{A}^{op}$, is exactly the same as the functor $\mathcal{A}(-,a)=H_a$.