$\def\R{\mathbb R}$$\def\aha{{1/2}}$$\def\maha{{-1/2}}$
Edit. Now it looks correct.
I can prove that $x_t$ grows at most like $t$ for $t\geq 1$, up to a multiplicative constant.
For simplicity, assume that $\|f\|_{L^\infty}\leq 1$. Call
$$ 0<\varepsilon:=\min_{x\in[1,2]} f(x). $$
Now write $u$ as a convolution with the heat kernel (I will use the kernel for the equation $u_t=u_{xx}$ and drop the factor $1/2$, as this amounts simply to a linear rescaling of the time variable).
$$ u(t,x)=\int_\R f(y)\frac{1}{\sqrt{4\pi t}}e^{-\frac{(x-y)^2}{4t}}dy=\int_\R f(t^\aha z)\frac{1}{\sqrt{4\pi}}e^{-\frac{(t^\maha x-y)^2}{4}}dz. $$
Call $\Phi(x):=\int_{-\infty}^x\frac{1}{4\pi}e^{-z^2/4}dz.$
We split the above integral in two pieces and we estimate both from below:
$$ u(t,x)\geq -\int_{-\infty}^0 \frac{1}{\sqrt{4\pi}}e^{-\frac{(t^\maha x-y)^2}{4}} dz + \varepsilon\int_{t^\maha}^{2t^\maha} \frac{1}{\sqrt{4\pi}}e^{-\frac{(t^\maha x-y)^2}{4}} dz=$$
$$ = -\Phi(-t^\maha x) + \varepsilon\left(\Phi(t^\maha(2-x))-\Phi(t^\maha(1-x))\right). $$
Now we use the estimate on $\Phi$ (which can be checked by differentiating both sides)
$$ \Phi(z)\leq \frac{1}{\sqrt{4\pi}}\frac{4}{|z|+1}e^{-z^2/4},\quad\forall z\leq 0 $$
and Lagrange formula to show that for $x\geq 0$ and some $g\in[1,2]$ we have
$$ u(t,x)\geq \varepsilon\frac{1}{\sqrt{4\pi t}}e^{-\frac{(x-g)^2}{4t}}-\frac{1}{\sqrt{4\pi}}\frac{4}{|x/\sqrt t|+1}e^{-x^2/4t}\geq $$
$$ \geq \varepsilon\frac{1}{\sqrt{4\pi t}}e^{-\frac{(x-1)^2}{4t}}-\frac{1}{\sqrt{4\pi}}\frac{4}{|x/\sqrt t|+1}e^{-x^2/4t}= $$
$$ =\frac{1}{\sqrt{4\pi t}}e^{-\frac{x^2}{4t}}\left(e^{\log(\varepsilon) +\frac{x}{2t}-\frac{1}{4t}}-e^{\log(4)-\log(\frac{x}{t}+\frac{1}{\sqrt t}) }\right).$$
Simplifying, one obtains that for non-negative $x$, $u(t,x)$ is positive whenever
$$ \frac{x}{t}>h\left(\frac{1}{2\sqrt t}+\frac{1}{4t}-\log(\varepsilon/4)\right)-t^\maha,$$
where $h=g^{-1}$ is the inverse of the function $g(z)=\log(z)+z/2$.
After further estimates on $h$, one can show that $u(t,x)>0$ whenever $t\geq 1$ and
$$ x>Ct$$
for some constant $C$ depending on $\varepsilon$ ($C=-\log(\varepsilon/20)$ should work). By symmetry, a similar bound holds from below, which implies that
$$ |x_t|\leq Ct $$
for all $t\geq 1$.
Combined with the asymptotic formula for small $t$, this should give the desired result.
I am not entirely sure every step of the proof is correct, but in any case this idea should work to establish the bound you want.
