What is geometrically the Pontryagin class?
Pontryagin's original definition for his classes was an obstruction cycle as follows:
On the $n$ dimensional manifold $M$ take $(n-2i) +2$ vector fields in general position, and consider the points $x$ where they span a subspace (in $T_xM$) of dimension less or equal to $n-2i$. The set of such points $x$ form a cycle of codimenion $4i$ in $M.$ The dual cohomology class is $p_i(M).$
This definition might differ from the today accepted definition through Chern classes (as in the book by Milnor-Stasheff) by a second order class.
Some fractional Pontrjagin classes are obstructions to higher analogues of orientations/spin structures.
For example, a spin vector bundle $E \longrightarrow X$ admits a string structure if $\frac{1}{2}p_1(E) = 0$. In other words, a spin structure on $E$ determines a class $\lambda = \frac{1}{2} p_1(E) \in H^4(X; \mathbb{Z})$ such that $2\lambda = p_1(E)$, and this fractional first Pontrjagin class $\lambda$ is the obstruction to the existence of a string structure on $E$.
Similarly, if we go to the next nontrivial step on the Whitehead tower, we can try to define a so-called fivebrane structure on a string vector bundle $E \longrightarrow X$. In this case, the obstruction to the string vector bundle $E \longrightarrow X$ admitting a fivebrane structure is the fractional second Pontrjagin class $\frac{1}{6}p_2(E)$.
I don't think that $p_1$ distinguishes the tangent bundles of exotic $4$-spheres (if any). On an oriented smooth $4$-manifold $M$ Hirzebruch signature formula states that
$${\rm sign}(M)=\frac{1}{3}\int_M p_1(TM).$$
The signature of any homology $4$-sphere is zero since there is no homology in the $4$-th dimension.
There is one stupid way in which $p_1$ describes an obstruction, because $p_1$ is the $2$-nd Chern class of the complexification, and Chern classes have obstruction-theoretic descriptions.
The first Pontryagin class of a $4$-manifold $M$ appears in a nice integral formula of MacPherson and it involves the singularities of generic maps $M\to \mathbb{R}^4$. (I do not remember the reference at this moment.)