After reading the statement of the Yoneda lemma (Theorem 4.2.1), I understand that it states that the functor $[\mathscr A^{op},\mathbf{Set}](H_\bullet,-)$ is naturally isomorphic to the functor $-(\bullet)$. That is, $[\mathscr A^{op},\mathbf{Set}](H_A,X)\cong X(A)$ naturally in $(A,X)$. Leinster (p.95) then proceeds to expain what the domain and codomain of those functors are (and to decompose them as composite functors):
I wonder how he did that? For example let's take the functor $[\mathscr A^{op},\mathbf{Set}](H_\bullet,-)$. Fix the second argument (to justify the first factors in the above composition). As far as I understand, for a fixed $X$, the functor $[\mathscr A^{op},\mathbf{Set}](H_\bullet,X)$ is the composition of $H_\bullet$ with $[\mathscr A^{op},\mathbf{Set}](-,X)$. By definition (p.90), $$H_\bullet:\mathscr A\to[\mathscr A^{op},\mathbf{Set}].$$
But in Leinster's composition the first component of $H_\bullet ^{op}\times 1$ is $H_\bullet^{op}$, not $H_\bullet$. How did he find out that that he should take $H_\bullet^{op}$ as the first component (and not $H_\bullet$)?
One explanation I see is that he tried $H_\bullet$ first, but it turned out to be wrong because the domain of the next arrow, namely $[\mathscr A^{op},\mathbf{Set}](-,X)$, is by definition $[\mathscr A^{op},\mathbf{Set}]^{op}$, which is different from the codomain of the previous arrow $H_\bullet$ (which is $[\mathscr A^{op},\mathbf{Set}]$). So he decided to replace $H_\bullet$ by $H_\bullet^{op}$ to make sense of composition.
I don't like my explanation because it's based on trial and error, and I don't think that's how Leinter figured out that the first component must be $H_\bullet^{op}$. How did he actually find this out?
Your explanation is basically correct, but it's not quite right to say it's by trial and error. If you're already clear on what the Yoneda lemma is saying, it's really the only option.
Leinster knows he's ramping up to prove the Yoneda lemma, so he needs a functor with the same signature as $ev:\mathscr{A}^{op}\times[\mathscr{A}^{op},\mathbf{Set}]\to\mathbf{Set}$. Further, he knows this functor is going to plug, for any $A\in\mathscr{A}$, $H_A$ into the first argument of $[\mathscr{A}^{op},\mathbf{Set}](-,-):[\mathscr{A}^{op},\mathbf{Set}]^{op}\times [\mathscr{A}^{op},\mathbf{Set}]\to \mathbf{Set}$ by virtue of what the Yoneda lemma is supposed to say. But you can't plug in $H_\bullet$ for that first argument; it doesn't have the right codomain.
The very silly thing about opposite functors is that the opposite of $F:\mathcal{C}\to\mathcal{D}$---that is, $F^{op}:\mathcal{C}^{op}\to\mathcal{D}^{op}$---acts exactly the same on objects and on morphisms as $F$. So $H_\bullet^{op}$ is really "the same thing" as $H_\bullet$, but it has the right codomain to go into the first argument of the Hom-functor, and the right domain to put the whole composite in the same functor category as the evaluation functor. It's not about using one instead of the other as much as it is realizing that one is the other.