The remarks on the distinction between function and relation are now correct, however the way the proof runs, relations are not involoved, only functions. I'll say a few things about your last paragraph:
"Thus, I don't apprehend the block quoted sentence in the proof of
Theorem 13.7. How can x NOT be an element of f(x)? By Definition 12.1,
isn't every x∈A always mapped by f?"
In one sense, it is usually the case for functions that $x$ is not an element of $f(x)$. For example given $f(x)=x^2$ we would say that $2$ is not an element of $f(2)=4$, since under a simple interpretation numbers are not regarded as sets. Even in the case $f(1)=1$ we wouldn't say that $1 \in 1$, again since in a simple interpretation $1$ is not a set.
[Note: there is a "logic" interpretation of nonnegative integers in which each integer is the set of its predecessors, so that e.g. $4=\{0,1,2,3\}.$ With this view, one does have $x \in f(x)=x^2$ when $x=2$, since $2$ is a predecessor of $4$. But this logic interpretation would say for example that $1$ is not a member of $f(1)=1$ since $1=\{0\}$ in this "logic" interpretaion. A restriction of this logic interpretation is that it doesn't associate sets to real numbers other than nonnegative integers.]
So if you have, as in this proof, a context in which the question whether $x\in f(x)$ even makes sense to ask about specific $x$ in the domain, it at least should be the case that for $x$ in the domain the image of $x$ under $f$ should be a set. That is, if $A$ is the domain of $f$ then for each $a\in A$ the image $f(a)$ is a set, in the proof here the codomain of $f$ consists of the collection of subsets of $A$, and these are sets, so that asking if $x \in f(x)$ for specific $x \in A$ at least makes sense.
The final question of the quoted part above, "By Definition 12.1, isn't every $x \in A$ always mapped by $f$?" has the answer "yes", simply because to say $x$ is "mapped by $f$" only means that it is mapped to something. And since $A$ is the domain of $f$, naturally any $x$ in $A$ does get mapped to something.
The question is, in the setup of the proof where one has assumed the existence of a map $f:A \to P(A)$ which is onto, whether a given $x$ in $A$ happens to map to a subset $f(x)$ of $A$ for which it happens that $x \in f(x).$
Here's a simple example in which $A=\{1,2,3\}.$ Suppose $f(1)=\{1,2\}$, $f(2)=\{3\}$, and $f(3)=\{1\}.$ In this case, we have $1 \in f(1)$ but $2 \in f(2)$ and $3 \in f(3)$ are each false.
The point of the proof is that, no matter how one sets the map $f$ up, it cannot wind up being an onto map from $A$ to the power set $P(A)$. Other answers (and the text you quote) have already covered this. I'm only throwing these thoughts into an answer because you expressed in comments some remaining confusion about the situation, and hope this sheds some light on that.
Added material re. query in question supplement Jan 4.
I can't pinpoint why I still think $\color{#009900}{[ a \in A ] \in f(a)}$.
Don't the green parts of Velleman's definition above mean and
reveal $\color{#009900}{[ a \in A ] \in f(a)}$? What am I
misreading/misconceiving?
I think what you have is a confusion between the technical definition of $f$ as a collection of ordered pairs, as opposed to the notation $f(a)$, which refers to the (unique) second coordinate of the pair $(a,b) \in f$. From the definition, for each $a\in A$ there's a unique $b\in B$ for which $(a,b) \in f$, and the notation $f(a)$ is then used to denote that particular $b$. Note that $b$ is not in $f$, since $b$ is merely the second element of an ordered pair in $f$.
If one looks at the particular pair $(a,b)$ as it occurs in the technical definition of $f$ as a collection of ordered pairs, and replaces $b$ by $f(a)$, what we have is that
$$(a,f(a)) \in f.$$
But one must keep in mind that in this statement $f$ is a collection of ordered pairs. If say I have the function $f=\{(1,7),(2,4)\}$ we can say that $(1,f(1))=(1,7)\in f.$ But we cannot say that $7 \in f$ because $7$ is not one of the pairs $(1,7),(2,4)$ which are the only two things in $f$. All we can say is that $7$ is the second coordinate of one of the pairs in $f$.
For usual functions, in which domain and range are collections of numbers, this confusion would be unlikely. My guess is that, in the present case where the range is a collection of sets, one might be tempted to think $f(a) \in f$ always. But the same thing happens, e.g. if $(2,\{1,2,5\}) \in f,$ we still cannot say $f(2) \in f$, only that $f(2)$ is the second coordinate of a pair in $f$.
I hope this clears up the supplementary question.
Best Answer
$f(x) = x$ where $f : \Bbb R \to \Bbb Z$ is not a function.
A function $f : A \to B$ is a set of ordered pairs $(a,b)$ with $a \in A$ and $b \in B$ such that for every $a \in A$ there is exactly one $b \in B$ that pairs with that $a$, that is $f(a)$ exists and $f(a) = b$. In set builder form,
$$f = \{(a,b) : a \in A,\ f(a) = b \in B\}$$
Injectivity requires the reverse, that each $b$ only pairs with one $a$.
Now, for your proposed $f$, you have a bunch of $a \in \Bbb R$ that don't get mapped anywhere. What would that look like as an ordered pair? $(a,)$? This doesn't make sense. Every element of your domain must get mapped somewhere or else it isn't a function.