Does it mean that a differential form of degree k is a mapping $:M \to ∧^k(T^∗M)$?
Yes, if you understand this to hold for every point $p \in M$:
$$
d: p \to T^*_p M ∧ ...∧ T^*_p M
$$
i.e. a k-form $d$ maps every point in M to an element of the k-exterior product of the cotangent space at $p$.
If I am correct, "the kth exterior power of" should be followed by a vector space.
Strictly speaking, yes, but in differential geometry it is often understood that one talks about the construction on all vector spaces at all points p of M (e.g. tangent or cotangent spaces).
I was wondering if a "cotangent bundle" $T^∗M$ of a differentiable manifold M is a vector space?
No, there are no algebraic operations defined on points of a manifold, a priori.
At a point of M, how does the definition of a k-form above lead to an alternating multilinear map, i.e., how does an element of $∧^k(T^∗M)$ become an alternating multilinear $T_p M×⋯×T_p M \to R$, as stated in the following?
That's the part that John M already addressed in his answer. Here is an elementary example:
A 1-form eats a vector field, for example, in cartesian coordinates in $\mathbb{R}^n$, we have a global vector field
$$
\partial_x
$$
and a global 1-form
$$
d x
$$
, and for every point $p \in \mathbb{R}^n$ we have the relation
$$
d x_p (\partial_x)_p = 1
$$
(If you are not sure about this, you should try to plug in the definitions of the gadgets on the left side and see if you can compute the result.)
A two form would be, for example,
$$
d x \wedge d y
$$
which we can feed two vector fields $\partial_u, \partial_v$, but we know of the relation
$$
(d x \wedge d y)_p (\partial_u, \partial_v)_p = - (d x \wedge d y)_p (\partial_v, \partial_u )_p
$$
by definition of the wedge-product. Note that you first choose a base point p, then your two-form gives you an element in $T^*_p M \wedge T^*_p M$, which you can then feed two tangent vectors at $p$ to get a real number.
HTH.
Non vanishing (at, say, $p$) means that there is a vector $v$ in $T_pM$ such that $\alpha_p(v)\neq 0$. Similarly for the $k$-form, it means that there is a set of $k$ vectors such the form is nonzero if evaluated on these vectors.
Best Answer
In what one might call naive calculus, for each coordinate $x_i$, the differential $dx_i$ denotes a small (infinitesimal, even) change in $x_i$, so a covector $\sum_i a_i dx_i$ is an infinitesimal change.
On a manifold, coordinates are only local, not global, so we should also imagine that this covector sits at a particular point of $M$. If we want to have a covector varying smoothly at every point, this is a differential one-form.
If $f$ is a function, then the total differential of $f$ is the quantity $$df = \sum_i \dfrac{\partial f}{\partial x_i} dx_i, $$ which records how $f$ is changing, at each point.
A tangent vector at a point is a quantity $v = \sum_i a_i \dfrac{\partial}{\partial x_i}$; you should think of this as a vector pointing infinitesimally, based at whatever point we have in mind. You can measure the change of $f$ in the direction $v$ by pairing $df$ with $v$.
Summary: tangent vectors are infinitesimal directions based at a point, while covectors are measures of infinitesimal change. You can see how much of the change is occurring in a particular direction by pairing the covector with the vector.
Now higher degree differential forms are wedge products of $1$-forms. You can think of these as measuring infinitesimal pieces of oriented $p$-dimensional volumes. (Think about how the (oriented) volume of an oriented $p$-dimensional parallelapiped spanned by vectors $v_1,\ldots,v_p$ depends only on $v_1\wedge\cdots\wedge v_p$.)