I was reading these notes and I came up with a question about Definition 8.2.1 at page 283: what does it mean for a functor $F : C \to Ab$ to be representable? Is this definition related to that of a representable presheaf?
I presume that, by composing $F$ with the forgetful functor $G :Ab \to Set$, we could adapt the definition of representable presheaf to $F$ and conclude that $F$ is representable iff $G \circ F$ is representable as a presheaf, but I am not totally sure this is correct.
$\require{AMScd}$The keyword here is "enriched category"; what Neeman is leaving as understood is that every triangulated category is preadditive i.e. every hom-set is not only a set, but an abelian group, and every composition map $$ c: \hom(A,B)\times \hom(B,C) \to \hom(A,C)$$ is not only a function, but a $\mathbb Z$-bilinear map; this means that $c$ becomes a unique, well-defined map of abelian groups $$ c: \hom(A,B)\otimes \hom(B,C) \to \hom(A,C)$$ where the tensor product is understood over the ring $\mathbb Z$. The category of abelian groups here plays the role of a "base" where you can "enrich" your category (homs are groups now, they carry richer structure than mere sets).
Category theory can be cast in its entirety "enriched over set", and "enriched over another base"; yet, this is something you can overlook considering your final goal (which is, afaicu, understanding Brown representability?). Enriched category theory is both a very informative guiding principle and a very deep rabbit-hole, especially for the beginners. In case you need more information about that more general theory, you can ask, but I'd suggest you refrain from entering this specific hole: as you might have noticed, category theory opens very many Pandora's boxes very fast.
For the sake of understanding BRT, it is enough that you register that Yoneda lemma (the key concept in representability of functors) now exists in two forms, "strong" and "weak", specialised to the case where the "base of enrichment" (=the category each $\hom(A,B)$ in an object of) is $\bf Ab$: the weak form says that the underlying sets $FA$ and $\text{Nat}(\hom(-,A),F)$ of the two abelian groups are in bijection; the strong form (which takes a little bit more effort in general but is almost for free in the $\bf Ab$-case) says that they are also isomorphic as abelian groups.
As a general rule, you are rarely interested in knowing something about the set $FA$, when it is also a group, and mere bijections have little importance, you look for isomorphisms. A functor $F : \mathcal C^{op}\to \bf Ab$ is thus representable when there exists an object $R$ (as "r"epresentative) in $\mathcal C$ such that there is a natural isomorphism of abelian groups $\mathcal C(X,R)\cong FX$.
The word "natural" is what makes this result a little subtler than it seems because there can be a way (in fact, there is plenty) to force a bijection $\theta : \mathcal C(X,R)\overset{\sim}\to FX$ of underlying sets to become a group isomorphism: either side (say, $FX$) is a group, just transport this operation on the other side by posing $$ f+g := \theta(\theta^{-1} f + \theta^{-1} g)$$ In principle, though, there is no reason for this family of bijections $\theta_X : \mathcal C(X,R)\cong FX$ to be natural, in the sense that it makes the squares $$ \begin{CD} \mathcal C(X,R) @>\theta_X>> FX \\ @Vu^* VV @VVFuV \\ \mathcal C(Y,R) @>>\theta_Y> FY \end{CD} $$ all commute, for every morphism $u : Y \to X$ in $\mathcal C$. For this to be true, you have to build an isomorphism in a keener way, showing that the two objects $\mathcal C(-,R), F$ have -in a suitable sense- the same universal property as objects of the category of functors $\mathcal C^{op} \to \bf Ab$.