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.
If $x \mapsto \varphi_i \varphi_j^{-1}x$ are the transition functions of your atlas of $M$, then the transition functions of $TM$ are $$(x,y) \mapsto (\varphi_i \varphi_j^{-1}x, \text{Jac}(\varphi_i \varphi_j^{-1})y).$$
The Jacobian is a matrix of first partials, so because $\varphi_i \varphi_j^{-1}$ is a $C^k$ function, the function above is a $C^{k-1}$ function. It seems you are already comfortable with this point.
Now the atlas on the cotangent bundle has transition functions given by $$(x,y) \mapsto (\varphi_i \varphi_j^{-1} x, \left(\text{Jac}\left(\varphi_i \varphi_j)^{-1}\right)^T\right)y).$$ The inversion and transposition operations are smooth, so again these transition functions are $C^{k-1}$.
Best Answer
Hint: Denote the canonical projection by $p:\land^2M\to M$, then for each $U$, take $p^{-1}(U)\cong U\times A^2(T_pM)$.