Deforming a section to a section without zeros Let $M$ be an oriented manifold of dimension $n$. Suppose furthermore that $E$ is an oriented vector bundle of rank $n-1$ over $M$. Let $s$ be a section of $E$ transversal to the zero section in $E$. Thus the zero set $Z$ of $s$ is a union of embedded circles in $M$ (we assume that the zero set lies in the interior of $M$ in case $M$ has non-empty boundary)
Under what conditions (if any) is it possible to deform $s$ such that the deformed section $s'$ has no zeros anymore?
Note that, in case the rank of $E$ is equal to the rank of $M$, then this is always possible provided the Euler class of $E$ is zero.
 A: This is a classical problem in obstruction theory. There is indeed an extra obstruction living in $H^n(M,\pi_{n-1}S^{n-2})=H^n(M,\mathbb{F}_2)$. But this obstruction is defined only modulo an indeterminacy comming from $H^{n-2}(M,\pi_{n-2}S^{n-2})$. There are two dual ways to look at it, so let me try to explain both. We can consider the filtration on $M$ by skeleta, and then the picture is as follows: The condition that the Euler class vanishes exactly allows us to deform the section to be non-zero on the $n-2$-skeleton. Indeed, the Euler class can be defined by the cocycle that encode the mapping degrees of attaching maps of $n-1$-cells in some local trivialization of the bundle at each $n-1$-cell. By modifying it along the $n-2$-cells we can replace $e(E)$ by any cohomologous cocycle and so we can re-choose the cocycle to be $0$ onthe nose, allowing to deform the value in the interior of the $n-1$-cells into a non-vanishing one. 
Then, we can lift the resulting section on the $n-1$-skeleton to a section on $M$, and we get a section with zeros only inside the $n$-cells. 
We can ask if we can eliminate those zeros. choose a trivialization of the bundle on each cell, then the section on the $n-1$-skeleton give us a collection of maps $S^{n-1}\to \mathbb{R}^{n-1}-0\cong S^{n-2}$ and the collection of homotopy classes is now a function from $n$-cells to $\pi_{n-1} S^{n-2}$ which for $n>4$ is $\mathbb{F}_2$, represented by the stabilization of the Hopf fibration. So we get a cocycle that at each cell tells us if the map from the boundary is null homotopic or not. Note that the invariant in $\mathbb{F}_2$ corresponding to each cell is related to the cohomology operation $sq^2$ which detects the Hopf map.
Here is another, more algebraic way to understand this obstruction. We have a map $BSO_{n-1}\to K(\mathbb{Z},n-1)$ classifying the Euler class. DEnote the fiber by $F_1$, then one can compute the cohomology of $F_1$ with coefficients in $\mathbb{F}_2$ by the Leray spectral sequence and one find a cohomology class in degree $n$, as follows. We have a fibration 
$K(\mathbb{Z},n-2)\to F_1\to BO_{n-1}$ and by definition the fundamental class of $K(\mathbb{Z},n-2)$ has its transgression the Euler class $e$. On the level of chains, we choose a class $y$ in $C^*(F_1)$ such that $d(y)=\pi^*y$ for $\pi:F_1\to BO_{n-1}$. Then, since the cohomology of $K(\mathbb{Z},n-1)$ has the pattern $\mathbb{Z},0,\mathbb{F}_2,0,...$ starting from $n-2$, we see that there is a $\mathbb{F}_2$ component in $H^n(K(\mathbb{Z},n-2))$ and this is exactly the class that detects the Hopf map in $\pi_{n-1}S^{n-2}$.
Taken mod 2, this class in cohomology can be idetified with $sq^2(y)$ where $y$ is the foundamental class. 
Now, we know that $sq^2(w_{n-1})=w_{n-1}w_2$ in $H^*(BO_{n-1},\mathbb{F}_2)$ so in particular the class $sq^2(y)-w_2y$ servives the spectral sequence and gives a cohomology class in $H^n(F_1,\mathbb{F}_2)$, well defined modulo some ambiguities that ill ignore for now. 
If $E$ is a vector bundle with vanishing euler class, then choosing a way to write the euler class as a coboundary gives us a lift of the map $M\to BO_{n-1}$ along the map $\pi: F_1 \to BO_{n-1}$. Geometrically this is a choice of a surface with boundary the circles of zeros of a section precicely!
Then, the next obstruction is the pull-back along the lift of the class $sq^2(y)+w_2(y)$ in the cohomology of $F_1$.  
Now, at least informally (and there should be some secondary cohomology operations formalism that make this precise) the next obstruction is given by choosing a cochain $y$ with $d(y)=e(E)$ and then the obstruction is something like $sq^2(y)+w_2(E)\cup y \mod 2$. The problem here is that we need cohomology operation on the chain level so again there should be some secondary cohomology operations formalism that makes it precise. 
Now the ambiguity: Here its pretty straight foward what it is: We can change $y$ by a cocycle, so we need to consider the secondary obstruction modulo the classes of the form $sq^2(a)-w_2(E)a$ for $a\in H^{n-2}(M,\mathbb{Z})=H^{n-2}(M,\pi_{n-2}S^{n-2})$. 
Using Wu formula, in the case of oriented manifold we can re-write the term $sq^2(a)$ as $v_2 \cup a$ so the indetermenacy of the obstruction is the ideal generated by $v_2(M) + w_2(E)$ where $v_2$ is the second Wu class of the manifold $M$. 
It is interesting if there's a simple description in terms of the Zeros of the deformed section, but I don't know.          
Edit: In fact, one can recover the homotopy class of the section on the attaching map of a top cell using the zeros, in a way that was mensioned in the comments above. The zeros are now stably framed cicrcles inside the cell, and the homotopy class corresponding to the cell is the class of this stably framed circle. 
To see this, just take a point in the unit disc in $\mathbb{R}^{n-1}$, considered as the vector bundle trivialized at the cell, and drag it generically to zero. The fiber at the point in the unit disc is the class of the map from the boundary of the cell, and the preimage at the origin is the zero locus so they are equivalent.   
