Every square is a rectangle, but not every rectangle is a square. Similarly, every denumerable set is countable, but not every countable set is denumerable. If you want, think of "denumerable" as an abbreviation for "countable and infinite" (or think of "countable" as an abbreviation for "denumerable or finite").
Roughly speaking, Martin's axiom says that many statements that can be proven about countable sets by simple diagonalization arguments also hold for sets of all cardinalities below $2^{\aleph_0}$. Here is a typical example.
Two subsets of $\omega$ are called almost disjoint if their intersection is finite. Consider the following simple theorem.
Theorem: Let $S$ be a countable set of subsets of $\omega$ which are pairwise almost disjoint and such that the union of any finitely many elements of $S$ is coinfinite. Then there exists an infinite subset $A\subseteq\omega$ that is almost disjoint from every element of $S$.
Proof: Enumerate the elements of $S$ as $(A_n)_{n\in\omega}$. Define a sequence $(n_k)$ of distinct elements of $\omega$ as follows. Having defined $n_j$ for all $j<k$, let $n_k$ be some element of $\omega$ that is different than every such $n_j$ and is not in $A_j$ for all $j<k$. Such an element exists since $\bigcup_{j<k}A_j$ is coinfinite by hypothesis.
Now let $A=\{n_k\}$; since the $n_k$ are all distinct, $A$ is infinite. But for any $j$, $n_k\not\in A_j$ for all $k>j$, so $A\cap A_j$ is finite. Thus $A$ has the desired properties.
Now you might ask, what if $S$ is not countable? Well, clearly the theorem cannot remain true without some additional constraint on $S$, since you could take $S$ to be a maximal infinitely family of pairwise almost disjoint subsets of $\omega$ (which exists by Zorn's lemma, say). In particular, the theorem cannot be true for all $S$ of cardinality $2^{\aleph_0}$. But what if $S$ is uncountable, but $|S|<2^{\aleph_0}$? The proof of the theorem certainly wouldn't apply to $S$; trying to continue the construction of $(n_k)$ transfinitely runs into trouble already at $k=\omega$, since the $n_j$ you have chosen already could exhaust the entire complement of $\bigcup_{j<k} A_j$.
However, Martin's axiom implies that the theorem does apply to such $S$. In this sense, Martin's axiom says that sets of size less than $2^{\aleph_0}$ behave "as though they were countable" for these purposes.
Of course, this raises the question of exactly which "purposes" Martin's axiom applies to. The exact statement is quite technical (and probably won't mean much to you if you aren't familiar with forcing). It says that if $P$ is a poset satisfying the countable chain condition (given any uncountable subset $A\subseteq P$, there exist distinct $a,b\in A$ and $c\in P$ such that $c\leq a$ and $c\leq b$) and $S$ is a collection of fewer than $2^{\aleph_0}$ dense subsets of $P$ (a subset $D$ of $P$ is called dense if for each $a\in P$, there exists $b\in D$ such that $b\leq a$), then there exists a filter on $P$ which intersects every element of $S$. In the case that $S$ is countable, it is easy to construct such a filter by a diagonalization argument (pick a descending sequence of elements of each set in $S$ one by one, and take the filter they generate). So, this says that in a wide variety of situations (anything that can be encoded by finding a filter in a ccc poset meeting dense sets), sets of fewer than $2^{\aleph_0}$ conditions can be simultaneously satisfied in the same way that countable sets of conditions can be simultaneously satisfied by building an object to meet the conditions one by one.
Here's another example that may be more familiar. By the Baire category theorem, any intersection of countably many open dense subsets of $\mathbb{R}$ is dense. A consequence of Martin's axiom is that this is also true for intersections of fewer than $2^{\aleph_0}$ open dense subsets of $\mathbb{R}$.
Best Answer
Here is what the book says:
So the fact that $\aleph_0$ and $\mathcal{c}$ are unequal is already implied by $\aleph_0<\mathcal{c}$. What Theorem 8.15(e) tells us is that $\aleph_0$ and $\mathcal{c}$ are transfinite.