Here is an example from p.45 in Leinster:
As far as I understand, in the definition of $\mathbf{Ab}(G_{ab},A)\cong\mathbf{Grp}(G,U(A))$, $\bar \phi$ is not quite $\bar \phi$. It constructed as follows. Consider $\phi:G\to U(A)$. This group homomorphism is also a group homomorphism $\phi':G\to A$. Now by the universal property, there is $\bar {\phi'}$. So $\bar \phi$ should actually be $\bar{\phi'}$, right?
Also I don't understand where the author omitted the symbol $U$ in the diagram. The universal property says that for any abelian group $A$ and any homomorphism $\phi:G\to A$ there is a unique $\bar \phi$ that makes the diagram commute. So it doesn't look like $A$ should be replaced by $U(A)$. And $G$ is any group, abelian or not, so I don't think it should be $U(G)$.
Using the name $U$ is not the best here. Often $\iota$ or something similar will be used for inclusions. This need not be trivial. Nevertheless, for a typical definition of $\mathbf{Ab}$ and $\mathbf{Grp}$, $\iota$ is a pure inclusion, i.e. $\iota(A)=A$ and $\iota(\varphi)=\varphi$. The point, of course, is that the $\iota$ makes it clear that we're restricting to abelian groups. The "correct" diagram that Leinster alludes to (using $\iota$ instead of $U$) is: $$ \require{AMScd} \begin{CD} G @>\eta>> \iota(G_{ab}) \\ @| @VV\iota(\bar\varphi)V \\ G @>>\varphi> \iota(A) \end{CD}$$ (written as a square due to typographical limitations). Or to write out the formula, it states: $$\forall A\in\mathsf{Ob}(\mathbf{Ab}).\forall\varphi\in\mathbf{Grp}(G,\iota(A)).\exists!\bar\varphi\in\mathbf{Ab}(G_{ab},A).\varphi=\iota(\bar\varphi)\circ\eta$$
We could omit the $\iota$s given that $\iota(H)=H$ and $\iota(\psi)=\psi$ for all $H$ and $\psi$. However, there are several reasons not to. First, $\iota$ may not be trivial in which case we can't omit it. Second, it makes clearer the connection to the general case of a universal property/adjunction. Third, it's nice to have a name for the inclusion so we can say things like $(-)_{ab}\dashv\iota$. Fourth, having $\iota$ be explicit means we don't need to write off to the side "for all Abelian groups $A$". Instead, this information is made explicit in the diagram. $A$ (and $G$) are free to be anything that makes the diagram "type check". Fifth, generally it is pleasant to write expressions that "type check".
To illustrate the difference in explicitness, $$\begin{CD} G @>\eta>> G_{ab} \\ @| @VV\bar\varphi V \\ G @>>\varphi> H \end{CD}$$ seems like a completely sensible diagram when $H$ is non-abelian while $$\begin{CD} G @>\eta>> \iota(G_{ab}) \\ @| @VV\iota(\bar\varphi)V \\ G @>>\varphi> \iota(H) \end{CD}$$ is nonsense. Of course, you have to know what $\iota$ is, but at least you know that you need to know as opposed to some remark in some text that you could easily skim past.
I don't understand why you think we should have $\phi'$ instead of $\phi$. Either $\phi'$ is literally the same as $\phi$ in which case what's the problem? Or $\phi$ is different from $\phi'$ (because $\iota$ isn't trivial) in which case we definitely want $\phi$ not $\phi'$ which may not even exist.