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.
Take in $\Bbb R^2$, for instance, any tensor represented by a matrix which is not symmetric nor skew. Say, the matrix $$\begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix}$$fits the bill and provides $T = e_1\otimes e_1 + e_1\otimes e_2 = e_1\otimes (e_1+e_2)$, where $\{e_1,e_2\}$ is the standard basis.
Best Answer
Yes, you are correct assuming what you enumerated. It is very common, from my experience.
Note also that, for some authors, a differential k-form is just a section of $\Lambda^k M$, whereas a smooth/differentiable differential k-form is a smooth/differentiable section of $\Lambda^k M$. For practical reasons, once the reader gets comfortable with the naming, all the preceding adjectives are usually ommited for the sake of clarity or brevity and one just says k-form.