I think the simplest motivation is that of vector fields. We want to be able to assign a tangent vector to each point in the manifold, giving us a "field" of vectors on the manifold. That is, $F$ should be some map such that
$$ F(p)\in T_pM $$
for $p\in M$. So, what's wrong with just saying that? Well, nothing really, if you're only interested in the value of vector fields at a point. If you ever want to look beyond singular points, you need some structure connecting your different tangent spaces. For example, to look at continuity of $F$, we need the space of outputs of $F$ to have a topological structure; to look at differentiability, we need a differentiable structure.
So, we need some space which
- contains all the tangent vectors of $M$
- has the same "level" of structure that $M$ has
To satisfy (1), we simply glue together all the tangent spaces by taking a union. Since the tangent spaces are completely disconnected from one another, we can exemplify this fact by using a disjoint union
$$ TM = \coprod\limits_{p\in M} T_pM $$
The rest of the bundle structure is just there to "lift" the structure of $M$ onto $TM$. With that, we can now define vector fields as functions in the usual way
$$ F: M\to TM $$
and we are able to discuss continuity and differentiability to the extent that $M$ admits such properties. However, this definition isn't "complete" because it allows for e.g. attaching a tangent vector from $T_qM$ to $p$, which doesn't fit our idea of a vector field. This gives us another requirement,
- we need a way to determine which point a vector is tangent to
This is the bundle projection map, $\pi: TM\to M$, and so we can add the requirement on $F$ that $\pi(F(p)) = p$ everywhere.
Here, we created a bundle with a base manifold and a vector space at each point, but we could imagine a more general concept of bundle which just has some kind of space $B$ with some other kind of space $F_p B$ at each point $p\in B$. Even in this general setting, we can see the utility of attaching to each point of the base space some element of its attached space. We define a function
$$ \sigma: B\to FB = E $$
with $\pi(\sigma(p)) = p$ as a cross-section (or just section) of the total space $E$.
Applying this terminology to our original example, we can then reconstruct the more terse definition:
$$ \text{a vector field on } M \text{ is a section }\sigma\text{ of the tangent bundle } TM$$
The characteristic classes of a tangent bundle $TM$, treated as a manifold itself, are defined using the double tangent bundle $TTM$. Given the projection $\pi: TM \to M$, the double tangent bundle splits as $\pi^* TM \oplus \pi^* TM$, and we have $$w_1(TTM) = 2 w_1(\pi^* TM) \equiv 0,$$ so $TM$ is orientable as you've asserted.
Similarly, $$w_2(TTM) = 2w_2(\pi^* TM) + w_1(\pi^* TM)^2 \equiv \pi^* w_1(TM)^2,$$ so whether it vanishes depends on $w_1(TM)^2$ in the cohomology ring of $M$.
You can work out what the other Stiefel-Whitney classes $w_i(TTM)$ are, using the formula $w(TTM) = \pi^* w(TM)^2$, where $w = 1 + w_1 + w_2 + \cdots$ is the total Stiefel-Whitney class.
Best Answer
To specify a bundle $\tau_M$, you would have to specify two topological spaces $E$ and $B$, and a projection map $\pi$ between them, verifying some conditions along the way.
So Milnor does exactly this, he specifies $DM$ as the total space, he specifies $M$ as the base space, then $\pi$, and finally goes on to show that the conditions on $\pi$ hold.
$DM$ is the total space of the bundle $\tau_M$. They are not the same thing.