This is actually a question about *categories*; not only about the category that I mention here specifically. I only use category $\mathsf{Rel}$ as an example.

> 
> How to describe a morphism that belongs to $\mathsf{Rel}^{op}\left(B,A\right)=hom\_{\mathsf{Rel}^{op}}\left(B,A\right)$?
> 
> 
> As a triple $\left(A,R,B\right)$ or as a triple $\left(B,R,A\right)$
> where in both cases $R\subseteq A\times B$?
> 
> 
> 

[ACC](http://katmat.math.uni-bremen.de/acc/acc.pdf) tells me: "...Thus $\mathcal{A}$ and $\mathcal{A}^{op}$
have the same objects and, except for their direction, **the same morphisms**..."(3.5.).

It comes to $\mathcal{A}^{op}\left(B,A\right)=\mathcal{A}\left(A,B\right)$.
Denoting the compositions in $\mathcal{A}$ and $\mathcal{A}^{op}$
by $\circ$ and $\circ^{op}$ respectively we have: $f\circ^{op}g$
is defined iff $g\circ f$ defined and this as $f\circ^{op}g=g\circ f$.

CWM tells me: "... the arrows of $\mathcal{A}^{op}$ are arrows
$f^{op}$ in a **one-one correspondence** $f\mapsto f^{op}$ with the
arrows $f$ of $\mathcal{A}$..."(page 33).

Here we arrive at $f^{op}\circ^{op}g^{op}=(g\circ f)^{op}$

Category $\mathsf{Rel}$ has (small) sets as objects and relations as morphisms. If $R\subseteq A\times B$ then it can be recognized as
a morphism $A\rightarrow B$ in $\mathsf{Rel}$. The demand that homsets
are disjoint 'tells' us that formally we must do it with triples $\left(A,R,B\right)$.

ACC makes me tend to: $\left(A,R,B\right)\in\mathsf{Rel}\left(A,B\right)=\mathsf{Rel}^{op}\left(B,A\right)$.

CWI makes me tend to: $\left(B,R,A\right)=\left(A,R,B\right)^{op}\in\mathsf{Rel}^{op}\left(B,A\right)$.