This is question is regarding co-orientation of vector-space. We have following: 
I have worked with orientations of vector spaces and manifolds before, hence showing that sign of determinant of transformation matrix gives equivalence class is straightforward. However I don't understand what co-orientation means here:
Are we considering determinants of linear transformation on total space which take basis $\mathcal{B}$ of complement of $F$ to different basis $\mathcal{B'}$ while fixing basis $\mathcal{B_0}$ of F?
If so then why would this be any different from simply taking orientation of complement space of $F$?
Finally, to prove final claim about $E \cap F$ I proceed as follows: orientation of $E$ naturally induces orientation on $E \cap F$ i.e. $E \cap F$ is subspace of $E$ so we can choose basis of $E$ for in given equivalence class so that first $k$ vectors span $E \cap F$. Since $E$ and $F$ are transversal then any orientation of complement of $E \cap F$ consist of vectors which are either in $E$ or $F$. I am not sure how to progress from there.
PS: This Exercise is taken from Audin, M. & Damian, M. - Morse theory and Floer homology - book I am using to study Morse homology independently.
For simplicity lets assume $E$ is complementary to $F$. Choose a (positive/negative) orientation $B_0$ of $F$. The co-orientation of $E$ is now a choice of orientation $B$ of $E$ such that $B$ and $B_0$ adds up to a (positive/negative) basis $(B_0,B)$ of the total space.
If E and F are transversal, $(B_0,B)$ gives a positive orientation of the total space, the projection map from the total space to the quotient of the total space by the complement of the intersection of E and F take the orientation $(B_0,B)$ to a an orientation of E intersection F which is the co-orientation actually.
I hope this helps. You may want to verify it a well.