First, note that the relation depicted in Explanation 8.5.11.1.1 is not a surjective partial function, but it is an epimorphism in $\mathsf{Rel}$ by Proposition 8.5.11.1.3, the next proposition. Moreover, partial surjective functions are epimorphisms by Proposition 8.5.11.1.3. For the rest of the proposition, we proceed by showing:
Note that we have
\[ R_{!}(R^{-1}(b))\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ b'\in B\ \middle |\ R^{-1}(b')\cap R^{-1}(b)\neq \text{Ø}\right\} . \]
We now claim Item 1 and Item 2 are equivalent:
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Item 1$\implies $Item 2: We proceed in a few steps:
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Since $R$ is functional, $R^{-1}(b)$ has at most one element.
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Since $R$ is surjective, $R^{-1}(b)$ has at least one element.
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Thus, $R^{-1}(b)$ is a singleton.
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The set $R(R^{-1}(b))$ will then be precisely $\left\{ b\right\} $.
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Item 2$\implies $Item 1: We claim $R$ is functional and surjective.
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Functionality. The inclusion
\[ R_{!}(R^{-1}(b))\subset \left\{ b\right\} \]
implies that if $a\in R(b')$ and $a\in R(b)$, then $b=b'$. Thus $R$ must be functional.
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Surjectivity. The inclusion
\[ \left\{ b\right\} \subset R_{!}(R^{-1}(b)) \]
implies $R^{-1}(b)\neq \text{Ø}$, so $R$ must be surjective.
Since $R$ is functional and surjective, it is a surjective partial function.
We have
\begin{align*} [R_{!}\circ R^{-1}](U) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}R_{!}(R^{-1}(U))\\ & = R_{!}\left(R^{-1}\left(\bigcup _{u\in U}\left\{ u\right\} \right)\right)\\ & = R_{!}\left(\bigcup _{u\in U}R^{-1}(\left\{ u\right\} )\right)\\ & = \bigcup _{u\in U}R_{!}(R^{-1}(\left\{ u\right\} ))\\ & = \bigcup _{u\in U}\left\{ u\right\} \\ & = U \end{align*}
for each $U\in \mathcal{P}(B)$, where we have used:
Taking $U=\left\{ b\right\} $ gives $R_{!}(R^{-1}(b))=\left\{ b\right\} $.
We have
\begin{align*} R_{*}(R_{-1}(U)) & = \left\{ b\in B\ \middle |\ R(R^{-1}(b))\subset U\right\} \\ & = \left\{ b\in B\ \middle |\ \left\{ b\right\} \subset U\right\} \\ & = U. \end{align*}
Suppose that for each $U\in \mathcal{P}(B)$, we have $R_{*}(R_{-1}(U))=U$. We must show that $R$ is functional and surjective.
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Functionality: We show that if $b,b'\in R(a)$, then $b=b'$.
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Consider the singleton set $U=\left\{ b\right\} $. By the assumed identity, we have
\[ \left\{ b\right\} =\left\{ b\in B\ \middle |\ R(R^{-1}(b))\subset \left\{ b\right\} \right\} . \]
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Since $b$ is an element of the set on the left-hand side, it must satisfy the membership condition on the right-hand side. Thus, we have $R(R^{-1}(b))\subset \left\{ b\right\} $.
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By assumption, $b\in R(a)$, which implies $a\in R^{-1}(b)$.
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By assumption, we also have $b'\in R(a)$.
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Since $a\in R^{-1}(b)$, it follows that the image of $a$ is contained in the image of the set $R^{-1}(b)$, i.e., $R(a)\subset R(R^{-1}(b))$.
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Combining these steps, we have $b'\in R(a)\subset R(R^{-1}(b))$.
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As we established that $R(R^{-1}(b))\subset \left\{ b\right\} $, we must have $b'\in \left\{ b\right\} $.
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Therefore, $b'=b$, which shows $R$ to be functional.
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Surjectivity: We show that for each $b\in B$, the preimage set $R^{-1}(b)$ is non-empty.
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Consider the empty set $U=\text{Ø}$. By the assumed identity, we have
\[ \text{Ø}=\left\{ b\in B\ \middle |\ R(R^{-1}(b))\subset \text{Ø}\right\} . \]
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The identity thus states that there is no element $b\in B$ for which $R(R^{-1}(b))$ is the empty set.
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In other words, for each $b\in B$, we must have $R(R^{-1}(b))\neq \text{Ø}$.
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The image of a set $R(S)$ is empty iff the set $S$ is empty.
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Therefore, the condition $R(R^{-1}(b))\neq \text{Ø}$ is equivalent to the condition $R^{-1}(b)\neq \text{Ø}$.
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Thus, $R$ is surjective.
Since $R$ is both functional and surjective, it is a surjective partial function. This finishes the proof.
A relation $R\colon A\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}B$ will be corepresentably faithful in $\boldsymbol {\mathsf{Rel}}$ iff, for each $X\in \operatorname {\mathrm{Obj}}(\boldsymbol {\mathsf{Rel}})$, the functor
\[ R^{*}\colon \mathbf{Rel}(B,X)\to \mathbf{Rel}(A,X) \]
given by precomposition by $R$ is faithful. This happens iff the morphism
\[ R^{*}_{S,T}\colon \operatorname {\mathrm{Hom}}_{\mathbf{Rel}(B,X)}(S,T)\to \operatorname {\mathrm{Hom}}_{\mathbf{Rel}(A,X)}(S\mathbin {\diamond }R,T\mathbin {\diamond }R) \]
is injective for each $S,T\in \operatorname {\mathrm{Obj}}(\mathbf{Rel}(B,X))$.
However, since $\boldsymbol {\mathsf{Rel}}$ is locally posetal, the Hom-set $\operatorname {\mathrm{Hom}}_{\mathbf{Rel}(B,X)}(S,T)$ is either empty or a singleton, As a result, the map $R^{*}_{S,T}$ will necessarily be injective in either of these cases.
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