Riehl's Category in Context - Example 2.1.5 (xi), Representable Functors "The functor iso: Cat → Set that takes a small category to its set of isomorphisms
(pointing in a specified direction) is represented by the category I, with two objects
and exactly one morphism in each hom-set".
So, if A and B are isomorphic in a category C, then we include only $f$ in our set $iso(C)$ (and not the inverse of $f$, since we are chosing isomorphisms pointing in a specified direction). Then, doesn't it suffice to have exactly one morphism in $I(a,b)$, where $a,b$ are the objects of $I$, and no morphisms in $I(b,a)$?
 A: To answer this question, I think it helps to first look a bit more closely at the category $I$. This category is sometimes call the "walking isomorphism" or the "free isomorphism". There are differnt equivalent ways to define it but the most common is the one you listed: $I$ is the category that has two objects, $0$ and $1$, and exactly one arrow in each hom-set.
What does this definition have to do with isomorphism? First, since $I$ is a category, it needs to have identities $1_0\in Hom(0,0)$ and $1_1\in Hom(1,1)$. But, since there is exactly one arrow in each hom set, these are the only arrows in these hom sets. So $Hom(0,0)=\{1_0\}$ and $Hom(1,1)=\{1_1\}$. But there are a few more hom sets, each with exactly one arrow. So $Hom(0,1)=\{f\}$ and $Hom(1,0)=\{g\}$, where $f$ and $g$ are just arbitrary names. So the category looks like this:

Now, since $I$ is a category, it must also contain the composite $gf$. But, by looking at the domain and codomain of $f$ and $g$, we can see that $gf\in Hom(0,0)=\{1_0\}$. So we must have that $gf=1_0$. We can similarly derive that $fg=1_1$. Thus, we must have that $f$ (and $g$) is an isomorphism!
Now lets see why $iso \cong Cat(I,-)$. Take any functor $F : I \to C$, i.e., an element of $Cat(I,C)$. Well, functors preserve isomorphism. So $F(f)\in C$ must be an isomorphism, with inverse $F(g)$! So any $F\in Cat(I,C)$ determines an element $F(f)\in iso(C)$. Now take an element $h\in iso(C)$. Since $h: x\to y$ is an isomorphism, it must have an inverse $h^{-1}:y\to x$. We can now define a functor $I\to C$ by specifying $0\mapsto x$, $1\mapsto y$, $f\mapsto h$, and $g\mapsto h^{-1}$, and sending identities to identities. Due to the way $I$ is defined, this fully specifies our wanted functor. And indeed, it is easy to verify that this mapping satisfies the functor laws. Whats important, though, is that $h$ is an isomorphism. If $h$ was not, then the mapping we defined would not satisfy the composition laws, also violating the theorem that functors preserve isomorphism.
Now, this is when thinking of $h$ as the isomorphism with inverse $h^{-1}$. We can also think of $h^{-1}$ as the iso, with inverse $h$. This would determine a different functor $F'$, that sends $f\mapsto h^{-1}$. But this would be a different functor. Of course, anytime we have one of these functors, we must have the other. But this is fine since any time we have an isomorphism, we must have a second isomorphism, its inverse.
It is a good exercise to prove that the correspondence between $iso(C)$ and $Cat(I,C)$ really defines a natural isomorphism. But I hope that the idea behind why $I$ represents $iso$, and thus earns the name "walking isomorphism" is much more clear.
What would happen if, as you suggest, we only require a morphism $f:0\to 1$? Well, there is actually a category like this, called the walking arrow and often denoted $2$. The problem with this is that $f\in 2$ is not an isomorphism! Thus, there is nothing that guarantees $F(f)$ is an isomorphism, for $F:2\to C$. However, $2$ has an interesting property. Namely, $Ar \cong Cat(2,-)$, it represent the functor that takes a category to its set of arrows.
