A complex vector bundle and its dual bundle are in general not isomorphic. As your argument shows, if $E \longrightarrow M$ is a complex vector bundle of odd rank $n$, then $c_n(E) = -c_n(E^\ast)$ and $e(E_{\Bbb R}) = - e((E^\ast)_{\Bbb R})$, so that both $E \not\cong E^\ast$ and $E_{\Bbb R} \not\cong (E^\ast)_{\Bbb R}$ (if $c_n(E) \neq 0 \neq e(E)$).
Note that this is in contrast to the case of real vector bundles. A Euclidean metric on a real vector bundle $F \longrightarrow M$ gives an isomorphism
$$F \cong F^\ast = \mathrm{Hom}(F, \underline{\Bbb R}).$$
The difference in the complex case is that a Hermitian metric on a complex vector bundle $E \longrightarrow M$ gives an isomorphism
$$\bar{E} \cong E^\ast = \mathrm{Hom}(E, \underline{\Bbb C}),$$
where $\bar{E}$ is the conjugate bundle of $E$. The conjugate bundle is not necessarily isomorphic to the original bundle.
Here's a direct connection between the index of a vector field and the obstruction theory definition of the Euler class. Note, I'm using the definition of the obstruction class from Wikipedia, since you link to it in the question.
Let $M$ be an $(n+1)$-manifold, because the indexing is slightly easier.
Let $V$ be a vector field on $M$ with finitely many zeroes. Triangulate $M$ such that each zero is in the interior of a different $(n+1)$-simplex, oriented so that it agrees with the orientation on $M$.
Then $V|_{M^{[n]}}$ (meaning the restriction of $V$ to the $n$-skeleton of $M$) is nonvanishing, and thus defines a partial section of $T^1M$ on the $n$-skeleton. The obstruction cochain will be an $n$-cochain with values in $\pi_n(S^n)\cong \Bbb{Z}$ (note that the orientation determines the isomorphism of $\pi_n$ of the fiber, $S^n$ with $\Bbb{Z}$).
The value of the $(n+1)$-cochain on a particular $(n+1)$-simplex $\Delta$ is given by $V(\partial \Delta)\in \pi_n(p^{-1}(\Delta))\cong \pi_n(S^n)\cong \Bbb{Z}$, where $p:T^1M\to M$ is the projection.
Now it's worth going over the details of this computation here.
If $\Delta$ doesn't contain a $0$ of $V$,
then $V(\Delta)$ fills in $V(\partial \Delta)$, so $V(\partial\Delta)$ is $0$ in $\pi_n(p^{-1}(\Delta))$, and thus the final integer value is $0$.
On the other hand, if $\Delta$ does contain a $0$ of $V$, $x$, then
by definition, the index of $V$ at $x$ is the degree of a map from $S^n\to S^n$ where the first $S^n$ is the boundary sphere of a disk $D^{n+1}$ in $M$ on which $T^1M$ is trivializable, and $V$'s only zero is $x$, and the second $S^n$ is the standard sphere in $\Bbb{R}^n$. The map is the composite
$$ \partial D^{n+1} \xrightarrow{V} \partial D^{n+1}\times (\Bbb{R}^{n+1}\setminus\{0\}) \to S^n,$$
where the second map is the projection onto $\Bbb{R}^{n+1}\setminus\{ 0\}$ followed by normalization. Note that the first map depends on a choice of trivialization, which should be chosen to be compatible with the orientation.
For convenience, note that we can dispense with requiring $T^1M$ to be trivializable, since the degree is a homotopy invariant, so it suffices that $T^1M$ is trivializable up to homotopy, which is in fact the case for any choice $D^{n+1}$. In particular, it is true for $D^{n+1}=\Delta$. Additionally, the degree of the map is usually computed with either homology (or cohomology), but in our case, the degree can also be computed with homotopy groups, since for the sphere we have canonical isomorphisms $H_n(S^n)\cong \pi_n(S^n)$ by the Hurewicz theorem.
However, if we use this to translate things over, we see that the value of the obstruction cochain on $\Delta$ is the index of the vector field $V$ at $x$.
Now we're almost done.
The isomorphism of $H^{n+1}(M)$ with $\Bbb{Z}$ for a closed, connected, oriented manifold $M$ is given by evaluating an $(n+1)$-cochain at the fundamental class.
If we call the obstruction $(n+1)$-cochain $\delta$, then
the integer we want is
$$
\sum_{\Delta} \delta(\Delta) = \sum_{x} \operatorname{index}_x V.
$$
Best Answer
(part of comment above converted to answer)
Note that the Euler class is only defined in the case of an oriented bundle, so you are assuming your manifold to have an orientation. In that case, your argument is correct. As you noted, the Euler class is the one and only obstruction to finding a section of the sphere bundle of the tangent bundle, i.e. a nowhere-zero vector field on the manifold.