Which manifolds are parallelizable?
(Disclaimer. The following describes some cohomological conditions that formally answer the question. But they are 1) usually too hard to compute; 2) don't really explain what is the class of parallelizable manifolds geometrically.)
The problem of existence of a section is answered (well, in a sense) by obstruction theory. Namely, there is the first obstruction $o_1\in H^1(X,\pi_0(F))$ and there is a section on $sk_1(X)$ iff $o_1=0$; if $o_1=0$ each section on $sk_1(X)$ defines an obstruction $o_2\in H^2(X,\pi_1(F))$ and so on (and if all obstructions are trivial, the bundle has a section).
(Well, actually one should be careful with $H^1(X;\pi_0(F))$: in general, $\pi_0(F)$ is not a group, so this $H^1$ just doesn't make sense, and the story starts a step (or two, if $\pi_1(F)$ is not abelian) later. But in the cases we're interested in, $o_1$ is well-defined.)
In the case of frame bundle of $n$-dimensional vector bundle, the fiber is $O(n)$, so obstructions lie in groups $H^i(M,\pi_{i-1}O(n))$. In the stable range homotopy groups of orthogonal groups are given by Bott periodicity. If we're talking about tangent bundle, we care only about $\pi_{i-1}(O(n))$ for $i\leq n$ and for $i<n$ these groups lie in the stable range.
A (toy) example: for $S^3$ only nontrivial (reduced) cohomology group is $H^3(S^3)$; but $\pi_2 O(n)=0$ — so any vector bundle on $S^3$ is trivial (well, not the simplest proof of the fact, but still).
In case of vector bundles these obstructions can be also described more geometrically (in the spirit of characteristic classes theory).
- First obstruction $o_1\in H^1(M;\pi_0 O(n))=H^1(M;\mathbb Z/2\mathbb Z)$ is nothing else but $w_1$, the first Stiefel-Whitney class. It gives the obstruction to orientability — i.e. to reducing the structure group of the bundle from $O(n)$ to $SO(n)$.
- If the bundle is oriented, second obstruction $o_2\in H^2(M;\pi_1 O(n))=H^1(M;\mathbb Z/2\mathbb Z)$ is defined. It coincides with $w_2$ and gives the obstruction to the existence of a spin struction — i.e. to lifting structure group of the bundle from $SO(n)$ to its universal cover, $Spin(n)$.
- Next obstruction is defined for a spin bundle; $\pi_2O(n)=0$, so first non-trivial obstruction here is $o_4\in H^4(M;\pi_3 O(n))=H^4(M;\mathbb Z)$. In fact, it coincides with $\frac12p_1$ (where $p_1$ is the first Pontryagin class of oriented bundle). And it is the obstruction to lifting the structure group from $Spin(n)$ to (infinite-dimensional) topological group $String(n)$.
...And so on: the sequence of obstructions corresponds to the Postnikov tower $$ O(n)\gets SO(n)\gets Spin(n)\gets String(n)\gets FiveBrane(n)\gets... $$ (this is a kind of duality: one can think either about sequence of extensions of the section through the filtration of $M$ by skeleta, or about sequence of lifts through the Postnikov tower of $O(n)$).
Some references. Obstruction theory in general is discussed in the section 4.3 of Hatcher — but Hatcher uses the Postnikov-towers-approach (like in the second part of the answer), AFAIR. And more classical approach + obstruction-theoretic POV on characteristic classes is explained e.g. in section 12 of Milnor-Stasheff, I believe.
One more remark. As it is explained in the other answer, ordinary ("primary") characteristic classes can't answer the question, since they coincide for stably equivalent vector bundles. What obstruction theory gives is, in a sense, a theory of higher characteristic classes: secondary class (a priori) defined only if primary one is zero and so on.
Although characteristic classes give very nice necessary conditions for parallelizability and also some partial sufficient conditions, they are doomed to fail in general. Here is why. If the tangent bundle $TM$ of a manifold $M$ of dimension $n$ is stably trivial i.e. has the property that $TM\oplus \theta^r\simeq \theta^{n+r}$ for some integer $r$ (where $\theta $ is the rank one trivial bundle) , then all characteristic classes of $TM$ will be trivial (if they satisfy Whitney's axiom) . But there is no reason that $TM$ itself should be trivial.
For example the tangent bundle to any sphere $S^n$ is trivially (!) stably trivial [add the trivial normal bundle to get the trivial rank $n+1$ vector bundle] but only $S^1,S^3$ and $S^7$ are parallelizable.
An impressive theorem of Adams is that the maximum number of independent vector fields on $S^n$ or $\mathbb P^n$ is $8a+2^b-1$ where $n+1=k.2^{4a+b}$ with $k$ odd, $a\geq0$ and
$0\leq b\leq 3$ .
Some positive results
a) Steenrod has proved that every orientable 3-manifold is parallelizable.
b) Forster has proved some incredibly strong results on the analytic parallelizability (which is of course much stronger than differentiable parallelizability) of Stein manifolds.
For example every analytic submanifold of codimension one (=hypersurface) of $\mathbb C^n$ is analytically parallelizable. And a codimension two analytic submanifold of $X\subset \mathbb C^n$ is analytically parallelizable if and only if the first Chern class of its tangent bundle is zero: $c_1(TX)=0$
The real analogues of Forster's results are of course completely false: every sphere $S^n$ is a hypersurface of $\mathbb R^{n+1}$ and a submanifold of codimension 2 of $\mathbb R^{n+2}$ with vanishing Stiefel-Whitney classes (since its tangent bundle is stably trivial). However it is not parallelizable if $n\neq 1,3,7$ , as already mentioned.