let $\xi$ be an arbitrary vector bundle. Is $\xi\otimes\xi$ always orientable? Let $\xi=(E,p,B)$ be a line bundle (not nec. orientable). Then the tensor product $\xi\otimes \xi$ is orientable. 
I obtain this by choosing $b\in U\cap V$, $U,V$ open in $B$ such that $\phi_U:p^{-1}(U)\longrightarrow U\times \mathbb{K}^n$,$\phi_V:p^{-1}(V)\longrightarrow V\times \mathbb{K}^n$. $(b,x)\in E$. Then $\phi_U(b,x)=(b,u)$, $u\in \mathbb{K}^n$,  $\phi_V(b,x)=(b,v)$, $v\in \mathbb{K}^n$. Then $\phi_V\phi_U^{-1}(b,u)=(b,v)$, $v=\lambda(b)u$. Thus for any $(b,x_1\otimes x_2)\in F_b(\xi_1\otimes \xi_2)$,
$\phi_V\phi_U^{-1}(b,u_1\otimes u_2)=(b,v_1\otimes v_2)=(b,\lambda(b)u_1\otimes\lambda(b)u_2)=(b,\lambda(b)^2 u_1\otimes u_2)$ hence the transition map is a positive scalar multiple and the tensor product is oriented line bundle.
Is my way correct or not? Is there more clear way to prove this?
In general, let $\xi$ be an arbitrary vector bundle. Is $\xi\otimes\xi$ always orientable? How to prove? I am confused. Thanks. 
 A: George's answer is absolutely correct, but here is a more direct argument, which avoids characteristic classes and uses only linear algebra. First of all, each real rank $N$ vector bundle $p:E\to X$ is determined by a cocycle $\xi_{ij}: U_{ij}=U_i\cap U_j\to GL(N, {\mathbb R})$, where $U_k$'s form an open cover of $X$ which trivializes the bundle $p$. (These functions are just the transition maps between different bundle charts.) By definition, the bundle $p$ is orientable if and only if $det(\xi_{ij}(x))>0$ for all $i, j, x\in U_{ij}$. Suppose that we have two vector bundles $E\to X, F\to X$, where $X$ is paracompact; in particular, we can assume that the trivializing covers for both bundles are locally finite. Then, by intersecting their trivializing covers, we can assume that they have a common trivializing cover. It is a straightforward computation to check that if $E, F$ are vector bundles  over $X$ with cocycles $(\alpha_{ij})$ and $(\beta_{ij})$ then the cocycle for their tensor product $E\otimes F$ is
$$
\gamma_{ij}(x)= \alpha_{ij}(x)\otimes \beta_{ij}(x).
$$ 
Here $\otimes$ is the Kroneker tensor product of matrices, see e.g. here: Tensor product of matrices represents the tensor product of the corresponding linear maps. Observe furthermore, that if $A$ is an $m\times m$ matrix and $B$ is an $n\times n$ matrix then $det(A\otimes B)=(det(A))^n(det(B))^m$, see the same wikipedia page.
Therefore, if $\alpha_{ij}=\beta_{ij}$ as in OP's question, then the determinant of the tensor product is positive. Hence, $E\otimes E$ is orientable. The same applies to George's remark about tensor product of vector bundles of even rank. 
A: Since I made propaganda for Stiefel-Whitney classes three hours ago, let me show how easy it is to solve your problem with them:  
The first Stiefel-Whitney class of the tensor product of two vector bundles $\xi, \eta$ of ranks $r,s$ is $$ w_1(\xi\otimes  \eta)= rw_1(\eta)+s w_1(\xi)  \in H^1(B,\mathbb Z/2 )           $$ and so  in particular $ w_1(\xi\otimes  \xi)= rw_1(\xi)+r w_1(\xi)=0 \in H^1(B,\mathbb Z/2 )            $  [recall that  $H^1(B,\mathbb Z/2 )$ is a $\mathbb Z/2$-vector space].
Since $w_1(\xi\otimes  \xi)=0$, the vector bundle $\xi\otimes  \xi$ is orientable.
Edit: a bonus
The displayed formula for the first Stiefel-Whitney class of a tensor product immediately implies by the same reasoning that the tensor product of two vector bundles both of even rank is orientable, independently of whether the vector bundles themselves are orientable or not !
Aren't Stiefel- Whitney classes amazing ?
A: If the base space is paracompact, you can always pull back the vector bundle over another space such that the pull back bundle splits. Then the statement holds by functoriality of Stiefel Whitney classes and injectivity of the map. For general base space case, I think the proof might be a bit involved, and I do not have a proof at the moment. 
