There are some elementary observations. First of all, your equation $r=0$ allows you to eliminate the variable $y_4$. So the field $K$ is the same as the field $\mathbb{R}(y_1,y_2)$. Moreover, your quadratic form, $a_1u_1^2 + a_2u_2^2 + a_3u_3^2 - a_4u_4^2$ (where $a_4=1$) is not reduced. Multiply the equation by $y_2$, define $$v_1=y_2u_1,\ v_2 = u_2,\ v_3=y_2u_3,\ v_4=u_4.$$ Then your new defining polynomial is $f = b_1v_1^2 + b_2v_2^2 + b_3v_3^2 -b_4v_4^2 = 0$, where now $b_1$, $b_2$, $b_3$ and $b_4$ are actually integral elements in $\mathbb{R}[y_1,y_2]$, namely, $$b_1 = 1-y_1-y_2,\ b_2 = \frac{25}{16}(y_2^3+\frac{1}{5}(5y_1-9)y_2^2+\frac{4}{25}(y_1^2-5y_1+6)y_2 + \frac{4}{25}(y_1-1)), $$ $$b_3 = 1,\ b_4 = y_2.$$ Next, over the Zariski open $U=\text{Spec}\ \mathbb{R}[y_1,y_2][(b_1b_2b_3b_4)^{-1}]$, there is a finite, etale degree 2 morphism $\nu:V\to U$ and a relative Brauer-Severi variety $\pi:P_V\to V$ such that your quadric bundle over $U$ is the Weil restriction of $P_V$ via $\nu$. Thus existence of a $K$-rational point of your original quadric is equivalent to the triviality of the element $[P_V]$ in the Brauer group of $V$.
The Brauer group of a real quasi-projective surface has been studied, e.g., by Nikulin and Krasnov. First of all, there is the Brauer group of the corresponding complex surface, which has a local and a global part. If you want to prove non-existence, then the local part already may be sufficient.