How is this "Stiefel manifold"-like fiber bundle called? How to characterize it? Let $M$ be a smooth manifold of dimension $n$. Consider the space $V_k(M)$ of pairs $(x,\alpha)$ where $x \in M$ and $\alpha$ is a linear embedding $\mathbb{R}^k \hookrightarrow T_x M$ (or equivalently, $k$ independent vectors in $T_xM$). This is topologized as a subspace of $M \times \operatorname{Map}(\mathbb{R}^k, TM)$ and it fibers over $M$ with $(x,\alpha) \mapsto x$. The fiber is (homotopy equivalent to) the Stiefel manifold $V_k(\mathbb{R}^n) = O(n) / O(n-k)$.
A slight variation of this requires an oriented manifold $M$ and asks gives a subspace $V'_k(M) \subset V_k(M)$ where the $\alpha$ are required to preserve orientation. The fiber is then $SO(n) / SO(n-k)$. I'm most interested in the case $k = n$, if it makes any difference.
How is this construction called? What do we know about the resulting manifold (the total space of the bundle)?
I know that if $M$ is parallelized, then both bundles are trivial, respectively $V_k(M) \cong M \times V_k(\mathbb{R}^n)$ and $V_k(M) \cong M \times V'_k(\mathbb{R}^n)$. Simply take a trivialization $\phi : M \times \mathbb{R}^n \to TM$, then there is an obvious map $M \times V_k(\mathbb{R}^n) \to V_k(M)$, $(x,v_1,\dots,v_k) \mapsto (x,\phi(v_1),\dots,\phi(v_k))$. More generally like all $SO(n)$ bundles $V'_k(M)$ is characterized by the Euler class and the Pontrjagin classes has an Euler class and Pontrjagin classes – what is the precise relationship? Are they related to the classes of the tangent bundle of $TM$ itself? I was not even able to say what $V'_2(S^2)$ looks like .
 A: I don't know a general name for these, but for the case $k=n$, you've defined the frame bundle of a smooth manifold (or oriented frame bundle in your second type of construction). If I was pressed for a name, I would probably call it the "partial frame bundle" or maybe "$k$-frame bundle". The significance of these is that (just as with the Stiefel manifolds for $\Bbb R^n$), if you then quotient out by $GL(k)$ to get rid of the oriented basis, you obtain a bundle parameterizing $k$-dimensional subspaces of $T_pM$; and sections of this bundle are the same thing as rank $k$ distributions on $M$. (For instance, if I was to define what it meant for a distribution to be smooth, I would probably construct this bundle and say that it means that the corresponding section of this bundle is smooth.) By picking a metric, your spaces are fiber homotopy equivalent to the so-called orthonormal frame bundles. The geometry is usually more easily phrased in terms of these, so henceforth I'll assume we've picked a metric.
The characteristic classes of the frame bundle are the same as those of the tangent bundle. This is because they're classified by the exact same map $M \to BO(n)$; that is, they have the exact same transition functions. You recover the tangent bundle from the frame bundle by the fiber product $\text{Fr}(M) \times_{O(n)} \Bbb R^n$, using the canonical action of $O(n)$ on both $\text{Fr}(M)$ and $\Bbb R^n$.
To calculate $V_2'(S^2)$ - this is an $SO(2)$-bundle over $S^2$, defined by the same clutching function in $\pi_1 SO(2)$ as the tangent bundle; that is to say, defined by the map $z \mapsto z^2$ (using the identification of $SO(2)$ with $S^1$). One calculates with their favorite definition of the Euler class that the bundle corresponding to $n \in \pi_1 SO(2)$ has Euler class $n \in H^2(S^2;\Bbb Z)$, and hence that the Euler class of $V_2'(S^2)$ is 2. (Of course, this should surprise no-one, as the Euler characteristic of $S^2$ is 2.) Similarly one may calculate the rest of the characteristic classes. The total space of this bundle is $SO(3)$; one has an identification because of the transitive faithful action of $SO(3)$ on $V_2'(S^2)$. In this way one may identify the bundle with $SO(2) \hookrightarrow SO(3) \to S^2$. Actually, this same argument shows that the oriented frame bundle of any sphere is given by $SO(n) \hookrightarrow SO(n+1) \to S^n$. Similarly for the non-oriented one.
