Let $M$ be a given manifold and $\xi$ be a given $k$-dimensional vector bundle over $M$. How to determine whether the underlying real vector bundle of $\xi\otimes\mathbb{C}$, i.e. the Whitney sum $\xi\oplus\xi$, is trivial or not? I want to try different methods as many as possible.


My attempt: To prove $\xi\oplus\xi$ is not a trivial bundle, we only need to prove the Stiefel-Whitney class
$$
1\neq (w(\xi))^2=(1+\sum\_{i=1}^k w\_i(\xi))^2=1+\sum\_{i=1}^k (w\_i(\xi))^2,
$$
i.e. there exists $1\leq i\leq k$ such that 
$$
(w\_i(\xi))^2\neq 0.
$$
But even the cup-product of
$$
H^\*(M;\mathbb{Z}\_2)
$$
is known, I still do not know how to determine the element $w\_i(\xi)$?


Are there any other methods to try?



non-triviality of the underlying real vector bundle of the complexification of a real vector bundle
It is a bit unclear what the question is asking exactly, so let me try to give some examples what can happen to a vector bundle $E$ when taking $E\oplus E$ (for simplicity, let me call this Whitney doubling). 


Certainly, as already mentioned in the question, the non-vanishing of the Stiefel-Whitney classes is a sufficient condition for non-triviality of $E\oplus E$, and the Stiefel-Whitney classes of $E\oplus E$ are of the form $w\_{2i}(E\oplus E)=w\_i(E)^{\cup 2}$. 


Another possibility to show non-triviality of $E\oplus E$ would be using the Pontryagin classes which are classes in $H^{4k}(M,\mathbb{Z})$. We have $2p(E\oplus E)=2p(E)^{\cup 2}$, and non-vanishing of the Pontryagin classes of $E\oplus E$ will imply non-triviality. In favourable cases, this will be possible by computations in rational cohomology.




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However, real vector bundles are not determined by their characteristic classes in general. Generally, classification results can be obtained using obstruction theory (to the extent that homotopy groups of $O(n)$ can be computed). The general formalism is explained e.g. in Hatcher's book on algebraic topology. For $M$ a manifold, the obstruction to lifting a map $M\to BSO(n)^{(k)}$ along $BSO(n)^{(k+1)}\to BSO(n)^{(k)}$ lies in $H^{k+1}(M,\pi\_{k-1}BSO(n))$, and the possible lifts are classified by $H^k(M,\pi\_{k-1}BSO(n))$. So the vector bundle classification hinges on knowledge of cohomology with coefficients in homotopy groups. Moreover, knowing the induced morphism on homotopy groups $\pi\_k(O(n))\to\pi\_k(O(2n))$ allows to analyze what happens to vector bundles under Whitney doubling. The following examples can be obtained by obstruction theory: 


* As a very simple example, the morphism $O(1)\to O(2)$ is null-homotopic, hence every line bundle $L$ will have $L\oplus L$ trivial (generalizing Ben McKay's answer).
* For $O(2)\to O(4)$, the only nontrivial map induced on homotopy groups is the reduction mod 2 $\mathbb{Z}\cong\pi\_1(O(2))\to\pi\_1(O(4))\cong\mathbb{Z}/2\mathbb{Z}$. In particular, the Whitney doubling of a rank 2 bundle will be trivial iff the Euler class is even.
* To get some more interesting examples, recall the paper of Kervaire ( Non-parallelizability of the $n$-sphere for $n>7$. Proc.N.A.S. 44 (1958), 280-283.) 
Lemma 1 states that the induced morphism $\pi\_k(O(n))\to\pi\_k(O(2n))$ is twice the stabilization morphism for $k<n-1$ (where the stabilization morphism happens to be an isomorphism). In particular, the stable range lifting classes in the obstruction classification will just be multiplied by $2$ under the Whitney doubling.
Now let's use this to get examples of bundles $E$ with trivial Stiefel-Whitney classes such that $E\oplus E$ is stably non-trivial. Consider a manifold $M=S^5/(\mathbb{Z}/8\mathbb{Z})$ which has $H^4(M,\mathbb{Z})=\mathbb{Z}/8\mathbb{Z}$. Any such class $\sigma$ gives a rank $5$ bundle on $M$ whose Stiefel-Whitney $w\_4$ class will be the reduction mod $2$, and whose Pontryagin class is $2\sigma$. If we choose $\sigma$ to be twice a generator, we get a bundle with trivial Stiefel-Whitney classes and Pontryagin class $2$-torsion. Now employ the result of Kervaire, which states that the morphism $\pi\_3(SO(5))\to\pi\_3(SO(10))$ induced by Whitney doubling is multiplication by two. This means, that the Whitney doubling will have lifting class four times a generator of $H^4(M,\mathbb{Z})$. Hence the Whitney doubling is stably non-trivial, but has all characteristic classes trivial.
* Better examples exist in higher dimensions. The homomorphism $\pi\_{4k}(BO)\to\mathbb{Z}$ defining the $k$-th Pontryagin class is multiplication with $(2k-1)!$ if $k$ is even and multiplication by $2(2k-1)!$ if $k$ is odd (see the paper of Bott and Milnor on parallelizability of spheres). In particular, we can construct bundles over $S^{13}/(\mathbb{Z}/8\mathbb{Z})$ whose only lifting class is twice a generator of $H^{12}$. These bundles will have all characteristic classes trivial, but will have non-trivial Whitney doubling.


