Intuition for nontrivial fiber bundles in terms of sections I understand the notion of a nontrivial fiber bundle with fiber $F$ over a base manifold $B$, as defined in terms of the projection map $\pi$: for any sufficiently small region $U \subset B$, the preimage $\pi^{-1}(U)$ is homeomorphic to the product space $U \times F$, but the preimage $\pi^{-1}(B)$ itself (the total space) is not homeomorphic to $B \times F$. The Mobius strip is a standard example for visual intuition.
However, physicists like myself often think of a fiber bundle in terms of its sections rather than its projection map. Is there an equivalent definition of a nontrivial bundle formulated in terms of its sections $\sigma$ (the right-inverses of $\pi$)? I.e. a statement of the form "a fiber bundle is nontrivial iff (some section $\sigma$ has)/(all sections $\sigma$ have) property $X$"? If not, is there any intuition for what the sections of a nontrivial bundle "look like"? I know that a principle bundle is nontrivial iff it does not admit any global section, but I'm curious how things work for general fiber bundles.
 A: 
Is there an equivalent definition of a nontrivial bundle formulated in terms of its sections σσ (the right-inverses of ππ)? 

Yes.  See below.

I.e. a statement of the form "a fiber bundle is nontrivial iff (some section σσ has)/(all sections σσ have) property XX"?

No.  See below for why I technically said no.

If not, is there any intuition for what the sections of a nontrivial bundle "look like"?

Individual sections can "look like" anything.
Answer:
Each $\pi^{-1}(U)$ comes with a (not unique) homeomorphism: $t_i: \pi^{-1}(U_i) \to U_i \times F$.  Thus over each intersection, we obtain transition functions
$f_{ij}: U_{ij} \to Aut(F)$, which satisfy the cocycle condition:
$$f_{ij}f_{jk} = f_{ik}.$$
So your bundle is trivial iff your choices of transition functions $t_i$ could be altered so that $f_{ij} =1$ for all ${ij}$. In fancy language, your bundle's triviality is measured by the Cech Cohomology generated by these $f_{ij}$.
What does this have to do with sections?  Well sections could be used to create the trivializations.  So start with sections, $s_i$, build associated trivializations, $t_i$, then build associated transition functions $f_{ij}$, and then your bundle is non trivial if you could have used different sections $s'_i$ so that all of the transition functions were the identity.
So the reason why I said no to the second question I answered is that it is not a matter of talking about some section having a property or all sections having a property.  But rather it is about the collection of sections having a property.
A: This answer discusses smooth fiber bundles.
In a way, every fiber bundle with structure group $G$ can be represented by a principal $G$-bundle, where $G$ can be any Lie group. So, if you understand principal bundles, all I need to do is explain the above claim.
Let $\pi:E\to B$ be a fiber bundle with fiber $F$ and structure group $G$. This means that $G$ is a Lie subgroup of $\mathrm{diff}(F)$, the diffeomorphism group of $F$, and that we have the following data:
$1)$ An open covering of $B$ by the sets $\{U_\alpha\}_{\alpha\in I}$.
$2)$ A trivialization $\psi_\alpha:U_\alpha\times F\to \pi^{-1}(U_\alpha)$ for every $\alpha\in I$, such that all the induced transition maps $\varphi_{\alpha\beta}:U_\alpha\cap U_\beta\to \mathrm{diff}(F)$ admit values in $G$.
The associated principal $G$-bundle, $P\to B$, can be described, as a set, as follows. For a point $b\in B$, the fiber $P_b$ consists of all the diffeomorphisms $F\to E_b$ which agree with the given bundle data. Namely, let $\alpha\in I$ be such that $b\in U_\alpha$. So $\psi_\alpha|_b$ is a diffeomorphism $F\to E_b$, and the fiber of $P$ over $b$ is defined by 
$$
P_b:=\psi_\alpha|_b\cdot G.
$$
This is well-defined, as all transition maps are in $G$. 
We now describe a smooth structure for $P$. By construction, the set $P|_{U_\alpha}$ is isomorphic to the set $U_\alpha\times G$ (an isomorphism is given by the trivialization $\psi_\alpha$). Equip $P$ with the smooth structure that turns all those set isomorphisms into diffeomorphisms. By construction, there is a natural $G$ action on $P$, which turns $P$ into a principal $G$-bundle.
Let us think of vector bundles as examples for the above construction. A real vector bundle $E\to B$ of rank $k$ is a fiber bundle with fiber $\mathbb{R}^k$ and structure group $GL_k(\mathbb{R})$. The above procedure leads in this case to the frame bundle of $E$. If $E$ is equipped with metrics on the fibers, then it becomes a fiber bundle with structure group $O(k)$. Then, the procedure leads to the orthonormal frame bundle. Similarly, any additional structure on $E$ which can be described in terms of the structure group leads to a smaller subbundle of the frame bundle.
Edit: This is a continuation of the previous paragraph. Say we have a vector bundle $E\to B$, and we want to determine whether this is a trivial bundle or not. As explained above, the vector bundle $E$ is encoded by its frame bundle $P$, which is a principal $GL_k(\mathbb{R})$-bundle. We know that $P$ is trivial if and only if it admits a global section. This means that $E$ is trivial if and only if it admits a global frame. In a similar way, one can understand any fiber bundle by examining its associated principal bundle.
