Direct sum of non-orientable bundles is orientable? Let $M$ be a smooth manifold, and let $E_1,E_2$ be two non-orientable vector bundles over $M$. 
Is $E_1 \oplus E_2$ orientable?
I am sure there is an easy answer, but somehow my search didn't result with anything useful.

As Qiaochu says, the answer is no, in general. However, there is more to be said here.
An orientation on a vector bundle is a global section of the associated orientation bundle. So, a vector bundle is orientable if and only if its associated orientation bundle (which is a double cover) is trivial.
Now, an observation: Let $V$ and $V'$ be two real vector spaces. Then an orientation on $V\oplus V'$ is equivalent to a bijection $\mathrm{or}(V)\to\mathrm{or}(V')$, where $\mathrm{or}(\cdot)$ denotes the set of orientations. 
As follows from the above observation, an orientation on the vector bundle $E_1\oplus E_2$ is equivalent to a bundle-isomorphism $\mathrm{or}(E_1)\to\mathrm{or}(E_2)$. In other words, $E_1\oplus E_2$ is orientable if and only if the orientation bundles of $E_1$ and $E_2$ are isomorphic. This is usually not the case.
Of course, the answer may change if the topology of $M$ is simple in some way. For example, if $M$ is the circle, then it has exactly two different double covers. Hence, every two non-orientable vector bundles have the same orientation bundle, and consequently, their direct sum is orientable.
Edit: As an example, let $M$ be the twice punctured plane, $$M=\mathbb{C}\setminus\{0,1\}.$$ Set $$P_0:=\left.\left\{(x,y)\in\mathbb{C}\times M\right|x^2=y\right\},\quad P_1:=\{(x,y)\in\mathbb{C}\times M|x^2=y-1\}.$$ So $P_0$ and $P_1$ are two different non-trivial double covers of $M$. Each one of them has an associated line bundle. Each of these line bundles is non-orientable, and so is their direct sum.
