Your diagram may be somewhat misleading in that the point $x,y$ seems to lie halfway between the tips of the two black vectors. If that were the case, then the ratio would trivially be $2$. Also, it's not clear (to me) what you're referring to as "perpendicular", since there are no right angles in the diagram.
For general position of $x,y$, you can find two vectors parallel to the two black vectors that add up to $(x,y)$ by solving a linear system of equations:
$$a\vec v+b\vec w=\vec x\;,$$
where $\vec v$ and $\vec w$ are the two black vectors, $\vec x=(x,y)$, and $a$ and $b$ are parameters to be determined by solving the resulting $2\times2$ system of equations. The two desired vectors are then $a\vec v$ and $b\vec w$.
[Edit in response to the comment:]
If we denote the components of the black vectors by $(x_1,y_1)$ and $(x_2,y_2)$, respectively, the above linear system of equations becomes
This can be solved explicitly using Cramer's rule:
Then the components of the two red vectors are $(ax_1,ay_1)$ and $(bx_2,by_2)$, respectively (where $a$ and $b$ are determined as above).