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.
For a beginner just starting to come to grips with these ideas, I think the most useful answer is this:
Except in one special situation (described below), there is
essentially no relationship between the exterior derivative of a
differential form and the differential (or pushforward) of a smooth
map between manifolds, other than the facts that they are both
computed locally by taking derivatives and are both commonly
denoted by the symbol $d$.
Differential geometry is loaded with notation, and sometimes we just run out of letters, so we have to overload a symbol by interpreting it in different ways in different situations. The fact that two things are represented by the same symbol doesn't always mean that they're "the same" in any deep sense.
The one situation in which the two concepts are directly related is for a smooth map $f\colon M\to\mathbb R$. In this case, we can consider $f$ either as a smooth map between manifolds or as a $0$-form. Considering it as a smooth map, for each $x\in M$, the pushforward is a linear map $df_x\colon T_xM\to T_{f(x)}\mathbb R$. Considering it as a $0$-form, its differential $df$ is a $1$-form, which means that for each $x\in M$ we have a linear functional $df_x\colon T_xM\to \mathbb R$. The link between the two is the fact that, because $\mathbb R$ is a vector space, there's a canonical identification $T_{f(x)}\mathbb R\cong\mathbb R$, and under that identification these two versions of $df_x$ are exactly the same map.
The excellent answer by @user86418 explains a sophisticated context in which both pushforwards and exterior derivatives can be viewed as special cases of a more general construction; but that's a context I wouldn't recommend that a beginner spend much time trying to come to terms with.
Best Answer
Consider the case of vector fields (it is completely analogous for covector fields i.e. one-forms).
A smooth vector field $X$ on some smooth manifold $M$ is a smooth section of the tangent bundle $TM$, i.e. to every $p\in M$ we assign a vector $X_p\in T_pM$ varying smoothly with $p$. The vectors $X_p$ are operators on smooth functions on $M$. Given a smooth function $f\in C^\infty(M)$ we can apply the vector $X_p$ to $f$ to yield a real number $X_p(f)\in \mathbb R$.
In this way one can define the action of a vector field on a smooth function: $(X(f))_p:=X_p(f)$. So given a smooth vector field $X$ and a smooth function $f$, we get a new smooth function $X(f)\in C^\infty(M)$. In differential geometry it is now customary to identify the induced mapping $$ C^\infty(M)\to C^\infty(M), \ f\mapsto X(f) $$ with the vector field $X$ itself. Alternatively one has the induced mapping $$ \Gamma(T^*M)\to C^\infty(M), \ \omega\mapsto X(\omega), $$ where $X(\omega)$ is defined by $(X(\omega))(p):=\omega_p(X_p)$. The key fact about this mapping is (as noted in comments) that it is $C^\infty(M)$-linear.
For differential forms (and indeed general tensor fields) one can proceed in a very similar fashion. Let $\omega$ be a smooth $k$-form on $M$. Given smooth vector fields $X_1,\dots,X_k$ on $M$, we can define a smooth function $$ \omega(X_1,\dots,X_k)\colon M\to \mathbb R, \ p\mapsto \omega_p(X_1(p),\dots,X_k(p)). $$ Similarily to vector fields, this induces a mapping $$ \Gamma(TM)\times \dots\times \Gamma(TM)\to C^\infty(M), \ (X_1,\dots,X_k)\mapsto \omega(X_1,\dots,X_k). $$ It is again customary to identify the form $\omega$ with this mapping. This induced mapping will be $C^\infty(M)$-multilinear.