Why are order-k differential forms sections of the kth exterior power of the cotangent bundle? The question I ask is in the title. This should be quite well-known, and in fact probably I am going to get the response that it is the definition. To convey my confusion, I have to convey my understanding of what is a differential form and what is the contangent bundle. To simplify things, we assume that our whole setup is immersed in the Euclidean space. 
$1$. Differential forms.
We take the definitions from Rudin, Principles of Mathematical Analysis. This is for an open set in $\mathbb R^n$.
Suppose $E$ is an an open set in $\mathbb R^n$. A $k$-surface in $E$ is a differentiable mapping $\Phi$ from a compact subset $D \subset \mathbb R^k$ into $E$. $D$ is called the parameter domain of $\Phi$ consisting of points $\mathbf u = ( u_{i_1}, \cdots , u_{i_k} )$. 
A differential form of order $k \geq 1$ in $E$ is a function $\omega$, symbolically represented by the sum
$$\omega = \sum a_{i_1, \cdots , i_k}(\mathbf x) dx_{i_1}\wedge \cdots \wedge dx_{i_k}$$
where the indices $i_1, \cdots , i_k$ range independently from $1$ to $n$, and so that $\omega$ assigns to each $k$-surface $\Phi$ in $E$ a number$\omega(\Phi) = \int_\Phi \omega$ 
, according to the rule
$$\int_\Phi \omega = \int_D \sum a_{i_1, \cdots , i_k}(\Phi((\mathbf{u})) \frac{\partial ( x_{i_1}, \cdots , x_{i_k})}{\partial ( u_{i_1}, \cdots , u_{i_k})}d\mathbf u $$
where $D$ is the parameter domain of $\Phi$, and the functions $a_{i_1}, \cdots, a_{i_k}$ are assumed to be real and continuous in $D$. 
So in the above definition the differential $k$-form is a certain integral for functions on compact $k$-surfaces. Thus a differential form can be treated as a measure for the $k$-surfaces, which can be integrated.
$2$. Cotangent bundle
We take this from wikipedia.
Let $M \times M$ be the Cartesian product of $M$ with itself.  The diagonal mapping $\Delta$  sends a point $p$ in $M$ to the point $(p,p)$ of $M \times M$.  The image of $\Delta$ is called the diagonal.  Let $\mathcal{I}$ be the sheaf of germs of smooth functions on $M \times M$ which vanish on the diagonal.  Then the quotient sheaf $\mathcal{I}/\mathcal{I}^2$  consists of equivalence classes of functions which vanish on the diagonal modulo higher order terms.  The cotangent sheaf $\Omega$ is the pullback of this sheaf to $M$.
Now, Def 2: A differential form $k$-form $\omega$ is a section of $\wedge^k\ \Omega$.
Question.
We consider an open set in the Euclidean space and look at the two definitions. A priori, to my eyes, both appear to be different things. It needs to be proved that they are the same. Please help me out with a reference with the required proofs.
 A: You have to work a bit to get those two definitions to agree, but it is all standard lore in differential geometry.  Both of the references in the comments - Spivak and Madsen & Tornehave - are good and should have what you need, the latter a bit more useful in my opinion.  But I am writing this answer because in neither text (or virtually any other introductory differential geometry text) will you encounter explicitly your definition of the cotangent bundle, or for that matter words like "sheaf" and "germ".  Such notions are useful in differential geometry, but it is not so crucial to incorporate them into the foundations the way it is in algebraic geometry.
A definition that you will see in books (in some form) proceeds as follows.  Given a point $p$ in $M$, define the tangent space $T_p M$ to be the vector space of point derivations of $C^\infty(M)$ at $p$.  If $(x_1, \ldots, x_n): U \subseteq M \to \mathbb{R}^n$ is a local coordinate system near $p$ then the directional derivative derivations $\frac{\partial}{\partial x_i}|_{p}$ form a basis of $T_p M$.  Construct a vector bundle $TM$ whose fiber over a point $x$ in $M$ is $T_x M$, and form its dual $T^*M$.  This is the cotangent bundle, and it takes only some basic techniques with sheaves to prove that it is the same as what you defined.
The equivalence of this definition with your first definition is just a bunch of coordinate calculations, complicated by the fact that Rudin defines surfaces extrinsically (in contrast to the intrinsic definitions that you and I produced).  From the intrinsic point of view, a differential k-form is a smooth section of the bundle $\wedge^k T^*M$.  As above, a local coordinate system $(x_1 \ldots x_n)$ on an open neighborhood $U$ of $M$ yields a trivializing frame $\frac{\partial}{\partial x_1}, \ldots, \frac{\partial}{\partial x_n}$ for $TM$ over $U$, and the corresponding dual frame for $T^*M$ over $U$ is denoted by $dx_1 \ldots dx_n$.  Thus a trivializing frame for $\wedge^k T^*M$ is given by $\{ dx_{i_{1}} \wedge \ldots \wedge dx_{i_{k}}: 1 \leq i_{1} \leq \ldots \leq i_{k} \leq n\}$, which more or less explains the local formula in your question.  The details of your integral formula are explained by noting that Rudin is defining a $k$ form on $\mathbb{R}^n$ with its standard coordinate system and pulling it back to a $k$ form on the surface along the embedding of the surface in $\mathbb{R}^n$.  Thus it is necessary to sort out how an intrinsic differential form behaves under coordinate change (from $\mathbb{R}^n$ coordinates to coordinates on the surface), and the whole point of the theory is that they change in a way which makes integration coordinate invariant.
One last comment.  Rudin defines differential k-forms for k-dimensional surfaces in $\mathbb{R}^n$ which comes naturally equipped with a notion of integration.  Of course most interesting manifolds (e.g. the sphere) are not the images of compact domains in $\mathbb{R}^k$ embedded in $\mathbb{R}^n$; this is the right LOCAL picture, but not the right global picture.  So to properly define the integral of a k-form over a k-manifold, it is necessary to define the global integral in terms of patched-together local integrals via a partition of unity.  Most books set up the theory this way and prove that the integral doesn't depend on the choice of partition of unity, but no books explain why, for example, the intrinsic integral of an n-form $f dx_1 \wedge \ldots \wedge dx_n$ over an open set in $\mathbb{R}^n$ (involving partitions of unity) agrees with standard Lebesgue integral of $f$ over that open set.  I suspect that one would have to sort out this kind of issue in order to REALLY prove that your two notions of differential form agree.  It's possible to work this out on your own, but Brian Conrad sorted it out in his "How to compute integrals" handout here:
http://math.stanford.edu/~conrad/diffgeomPage/handouts.html
You might find some of the other handouts helpful too.
