I want to point out something potentially misleading about Marek's answer. The n-categories he mentions are not categories, but generalizations of them, so the question still remains, why do categories only form a 2-category, that is, why do people stop after categories, functors, and natural transformations? Why don't people define modifications of natural transformations?
I think it is good to realize that categories really are in an essential way only 2-categorical, if you want interesting higher morphisms you do need to define something like a higher category. One way to think about it is this: natural tranformations are basically homotopies. To make this precise, take I to be the category with two objects, 0 and 1, one morphism from 0 to 1 and the identity morphisms. Then, it is easy to check that to specify a natural tranformation between two functors F and G (both functors C → D) is the same as specifying a functor H : C × I → D which agrees with F on C × {0} and with G on C × {1}.
So then we could get higher morphisms by saying they are homotopies of homotopies, i.e., functors C × I × I with appropriate restrictions. This works, and we indeed get some definition of modification, but it is not interesting as it reduces to just a commuting square of natural transformations, i.e., it can be described simply in terms of the structure we already had.
This is similar to what happens for, say, groups: you can think of a group as a category with a single object where all the morphisms are invertible (the morphisms are the group elements and the composition law is the group product). Then group homomorphisms are simply functors. This makes it sound as if groups now magically have a higher sort of morphisms: natural tranformations between functors! And indeed they do, they are even useful in certain contexts, but they're not terribly interesting: a natural transformation between to group homomorphisms f and g is simply a group element y such that f(x) = y g(x) y -1 . Again, this is described in terms of things we already new about (the group element y and conjugation), and is not really a brand new concept.
No, your point is right. It's a bit mysterious why Mac Lane writes $F\mu c=\tau c$, as clearly the needed condition is $F\mu=\tau$, a condition which implies in particular that $FT_0=S$ and $FT_1=T$, using Mac Lane's notation, at least in the first edition. Mac Lane even writes the correct condition at the end of the paragraph.
Note that you reproduced some of the construction slightly incorrectly. $\mu$ is the function that assigns $(x,\uparrow)$ to each $x\in \mathcal C$, not to each $(x,\delta)$, because $\mu$ is a natural transformation with domain $\mathcal C$. Thus the claim you starred should also have been that $F(\mu(x))=\tau x$, not $F(\mu(x,\delta))=\tau x$.
Best Answer
There are two directions to this proof.
One direction is that given a functor $\varphi: \mathcal C \times 2 \to \mathcal D$, there is a corresponding natural transformation $\varphi(-, 0) \to \varphi(-, 1)$. $\varphi(-, 0)$ is a whole functor $\mathcal C \to \mathcal D$. The action on objects is obvious (simply evaluate $\varphi$ at the pair $(c, 0)$. If you haven't seen this before, the action on morphisms might not be obvious. Morphisms in $\mathcal C \times 2$ are defined to be pairs of morphisms in $\mathcal C$ and $2$, so a priori, $\varphi(f, 0)$ doesn't make any sense. However, it's typical with functors of multiple variables that an object is also shorthand for the identity at that object. That is, $\varphi(f, 0)$ is $\varphi(f, id_0): \varphi(c, 0) \to \varphi(c', 0)$.
Then, the natural transformation $\varphi(-, 0) \to \varphi(-, 1)$ is simply $\alpha_c := \varphi(c, \to)$, where $\to$ is the unique arrow $0 \to 1$ in $2$.
The other direction is that given a natural transformation $\alpha: \mathcal F \to \mathcal G$, there is a corresponding functor $\varphi: \mathcal C \times 2 \to \mathcal D$ such that $\varphi(-, 0) = \mathcal F$ and $\varphi(-, 1) = \mathcal G$. The behavior of $\varphi$ on objects is determined by the conditions that it's equal to the given functors at $0$ and $1$. For example, $\varphi(c, 0) = \mathcal F(c)$.
That leaves the action of $\varphi$ on morphisms. $\varphi(f, \to): \varphi(c, 0) \to \varphi(c', 1)$, i.e. $\mathcal F(c) \to \mathcal G(c')$. The natural choice is then the diagonal of the commutative diagram
$$ \require{AMScd} \begin{CD} \mathcal F(c) @>{\mathcal F(f)}>> \mathcal F(c')\\ @V{\alpha_c}VV @VV{\alpha_{c'}}V \\ \mathcal G(c) @>>{\mathcal G(f)}> \mathcal G(c') \end{CD} $$
Finally, one should really show that going one direction then the other leaves you where you left off. Once functorality of $\varphi$ and naturality of $\alpha$ are proven, that gives a bijection between functors of that certain form and natural transformations.