Suppose that $f$ is decreasing in $x$. Let $x(t)$, $y(t)$ be two solutions of the ode. Then
$$
\dot{x}-\dot{y}= f(x,t)-f(y,t).
$$

Multiplying both sides by $x-y$ we deduce

$$
(\dot{x}-\dot{y})(x-y) =\big(f(x,t)-f(y,t)\big)(x-y)\leq 0,
$$
where the last equality holds because $f$ is decreasing.

Hence
$$
\frac{1}{2}\frac{d}{dt}\big(x-y)^2\leq 0.
$$
Thus the function $t\mapsto \big( x(t)-y(t)\big)^2 $ is decreasing so
$$
\big(x(t)-y(t)\big)^2\leq \big( x(0)-y(0)\big)^2,\;\;\forall t\geq 0,
$$
i.e.,
$$
\Big(\Phi(x_0,t)-\Phi(y_0,t)\Big)^2\leq \Big(x_0-y_0\Big)^2,\;\;\forall t\geq 0.
$$
In other words, for $t\geq 0$, $\Phi(x,t)$ is Lipschitz in $x$ with Lipschitz constant $1$ if $f$ is decreasing.