Since the equation for $x$ does not involve $y$, you can solve it first as a single ODE:
$$\dot x = f(x), \quad x(t_0)=x_0\tag1$$
Since $f$ is globally Lipschitz, the Picard theorem implies that the solution of (1) exists for all times, and is unique.
Now that $x$ is a known function, the equation for $y$ turns into a linear ODE
$$\dot y = g(x)y, \quad y(t_0)=y_0\tag2$$
Here $g(x)$ is a continuous function of $t$, being the composition of two continuous functions. It follows that the solution of (2) exists for all times, and is unique.
Remark #1: "continuous and Lipschitz" is redundant. A Lipschitz function is automatically continuous.
Remark #2, concerning your meta question. The Tumbleweed badge is not completely useless. It serves as a catalog of overlooked questions, and thus draws some (small) amount of attention toward them -- which is better than nothing.