Are these vector bundles, trivial bundle? We identify the  vector  space tensor  product $\mathbb{R}^{m} \otimes \mathbb{R}^{n}$  with $\mathbb{R}^{mn}$
Let  $X$ be the  space  of  all  non zero simple tensors $X=\{a\otimes b \mid a\in \mathbb{R}^{n} \setminus \{0\}, \;b\in \mathbb{R}^{m} \setminus \{0\}\}$.
Let  $\pi:\mathbb{R}^{mn}\setminus \{0\}  \to  \mathbb{R}P^{(mn-1)} $  be  the  natural  projection. Put $PX=\pi (X)$
Is   the  tautological line  bundle  restricted to $PX$, a trivial bundle?
Is the  following  bundle  $(E,X,q))$  a trivial bundle  over $X$:
$E=\{(x\otimes y, T) \mid T:E_{x} \to E_{y} \;\; \text{is  a  linear  map }$  where $E_{x}= \{v\in \mathbb{R}^{n} \mid  v.x=0\} $
$X$  is  the  space simple  tensors  and $q$ is  the obvious projection.
 A: There is an evident map $\mathbb{R}P^{m-1}\times\mathbb{R}P^{n-1}\to PX$, which is easily seen to be an isomorphism.  (This is called the Segre embedding.)  If we pull back the tautological bundle of $PX$, we get the tensor product of the tautological bundle $M$ over $\mathbb{R}P^{m-1}$ and the tautological bundle $N$ over $\mathbb{R}P^{n-1}$.  Let $x$ and $y$ denote the first Stiefel-Whitney classes of $M$ and $N$, so the mod $2$ cohomology of $PX$ is $\mathbb{F}_2[x,y]/(x^m,y^n)$.  The first Stiefel-Whitney class of $M\otimes N$ is $x+y\neq 0$, so $M\otimes N$ is not trivial.  
The rest of your question is notationally unclear.  I will interpret it as follows: you want to consider the bundle $M^\perp$ over $\mathbb{R}P^{m-1}$, whose fibre at $\mathbb{R}x$ is the orthogonal complement to $x$.  You also want to consider the bundle $N^\perp$ over $\mathbb{R}P^{n-1}$, and the resulting bundle $E=\text{Hom}(M^\perp,N^\perp)$ over $PX$.  (You said $X$ but I am guessing that you meant $PX$.  If you pull back to $X$ then $E$ certainly becomes stably trivial, but not obviously trivial.  Different methods would be needed to address that.)  Now $M^\perp$ has an obvious inner product, so it is isomorphic to its dual, so $E$ can be identified with $M^\perp\otimes N^\perp$.  Using $M\oplus M^\perp=m$ and $N\oplus N^\perp=n$ we obtain an isomorphism
$$ E \oplus (m\otimes N) \oplus (M\otimes n) \simeq 
    mn \oplus (M\otimes N).
$$
Each of these bundles $V$ has a total Stiefel-Whitney polynomial $f_V(t)=\sum_iw_i(V)t^{\text{dim}(V)-i}$, and standard methods give
\begin{align*}
 f_{M\otimes n}(t) &= (t + x)^n \\
 f_{m\otimes N}(t) &= (t + y)^m \\
 f_{M\otimes N}(t) &= t+x+y 
\end{align*}
The above isomorphism therefore gives
$$ f_E(t) (t+y)^m (t+x)^n = t^{mn} (t+x+y). $$
From this you can calculate $f_E(t)$ and thus $w_i(E)$ for all $i$.  The details depend on the parity of various binomial coefficients which appear when you expand out $(t+x)^n$ and $(t+y)^m$.  However, you will always get 
$w_1(E)=nx+my+(x+y)=(n+1)x+(m+1)y$, so $E$ can only be trivial if $n$ and $m$ are odd.  However, in that case it works out that $w_2(E)$ always contains $xy$ (and possibly also $x^2$ and/or $y^2$), so again $E$ is nontrivial.
