These are important questions. When one starts to define connections, one usually imposes the Leibnitz condition on it. But I am proceeding too fast. First, let's recall what a connection is. Here we come to one very confusing point, at leatleast, it was for me, when I started learning differential geometry: There are various ways of defining connections. Just to name some of them:
Define the Levi-Civita connection on a (Riemannian) manifold. The Levi-Civita connection is an affine connection that preserves the metric structure, i.e., $\nabla g = 0$ and that is torsion-free, i.e., $\nabla_{X}Y-\nabla{Y}X = [Y, X]$, where $[. , .]$ denotes the Lie-bracket of two elements of $\Gamma(TM)$. Then one usually starts by showing that, by this definition, the Levi-Civita connection (hereafter called LC connection) is uniquely determined.
However, one can even generalize this approach by at first considering linear connections, then metric ones and then affine ones. After all that one can turn to the derivation of the LevLevi-Civita connection. Personally, I like this approach pretty much because it shows very well why the LC connection is so important. Namely, it eases computations for the Lie-derivative which normally does not require the notion of the LC connection at all. Also there are many fascinating mathematical objects, such as Killing fields which can be calculated (or at least defined) effectively, if one introduces the notion of the LC connection.
One can even start more generally, by considering fiber bundles. This appraochapproach was taken by the French mathematician Charles Ehresmann (1905-1979). This approach has one fundamental advantage: Generality, however, it lacks the easy-to-grasp-effect (I like to assign a definition this value if it has a very intuitive meaning). Interestingly, this approach does yield the LC connection as a special case when one reduces fiber bundles to vector bundles and considers the bundles $\pi : TM \longrightarrow M$ as a special case and requires the base manifold $M$ to be equipped with a symmetric, positive definite 2-times covariant tensor field, i.e., the metric tensor $(g_{ij})_{i,j}$.
Why am I writing all this? Simply to sum up what we already know and how beautifully, at least, in my opionionopinion, all these definitions harmonize.
If one has finally obtained a manifold that is equipped with a Riemannian metric, then one goes further and asks what the term orientation means for general Riemannian manifolds. Although for submanifolds of $\mathbb{R}^{n}$ one has averya very intuitive picture of what orientability means. We could expand this picture by making use of the Nash Embedding theorem, but it turns that this is not necessary.
Example: Consider for instance the so-called MobiusMöbius strip. It is a typical example of a non-orientable two-dimensional submanifold of the EucliedanEuclidean 3-space.
Namely, from the tensor product spaces can be obtained the k-th exterior cotangent space. I would like to refer you to the book Introduction to Kähler Manifolds which contains and excelltnexcellnt treatment of exactly this. Via the equivalence relation one then breaks down the Leibnitz rule to the k-h exterior power cotangent bundle. The rules is exactly what Matt has written above.
Let's come to question two: We know that the volume form is $SO(n)$ invariant, where $n=dim(M)$ Since we define the volume form $\omega := dx^{1} \wedge ... \wedge dx^{n}$$\omega := dx^{1} \wedge \dots \wedge dx^{n}$, with the $dx^[i}$$dx^{i}$ for $i=1,..,n$$i=1, \dots, n$ to be the product of the basis of the cotangent bundle, we can make the following thought. We
We know how to "differenciate""differentiate" the basis w.r.t to the LC connection. When we calculate all that, we finally obtain that the volume form is parallel. This was how I learned it. Of course one can use representation theory here which is, in my eyes, very interesting and also more elegant.
If there are questiosnquestions, don't hesitate to contact me.