$S^3$ is a $3$ -surface as $f(x_1,x_2,y_1,y_2)=x_1^2+x_2^2+y_1^2+y_2^2$ gives $f^{-1}(1)= S^3$ and for every point of $S^3$ is a regular point, i.e., $\nabla f\ne 0 $ at every point of $S^3$.
Since the tangent space to $S^3$ at $p$, denoted $T_pS^3$ equals $\nabla f(p)^\perp$. One can define $X_i(x_1,x_2,x_3,x_4)$ to lie in $\nabla f(p)^\perp, i=1,2,3$.
$\nabla f(p)^\perp$ is of dimension $3$ and hence has $3$ linearly independent vectors in it, but I'm having difficulty in finding these linearly independent vectors. Hit and trial gives: $X_1(x_1,x_2,y_1,y_3)= (y_1,y_2,-x_1,-x_2), X_2(x_1,x_2,y_1,y_3)=(-x_2,x_1,-y_2,y_1), X_3(x_1,x_2,y_1,y_3)=(-y_2,-y_1,x_2,x_1)$ but these are $X_i$'s are not linearly independent when $x_1=0$. How do I get $X_i$'s such that $X_i(*)$ are linearly independent for every $*\in S^3$?
Identify $S^3$ with the unit quaternions, which makes them a Lie group under quaternion multiplication. Calculate the tangent space to $1$, which is a $3$ dimensional real vector space, and pick a basis for this space. Now since you're dealing with a Lie group, the pushforward of this basis under the multiplication map $q: S^3 \to S^3$ allows you to extend this vector field smoothly to the whole sphere, and they will be linearly independent everywhere, since the multiplication map is a diffeomorphism.
I don't know what specific fields this will give you when you're done, since I haven't done this calculation in a long time (and also it depends on the choice of basis). If you're still stuck, I'll expand this answer to include a calculation later.