What you're seeing is correct. When mixing two chirps as you're doing, you will get a doubling of the frequency.
Consider two chirps with a bandwidth $B$ and pulse length $\tau$. We'll use the complex form to simplify the math
$$x_1(t) = e^{j{\pi}\frac{\beta}{\tau}t^2}$$
$$x_2(t) = e^{j({\pi}\frac{\beta}{\tau} + \phi)}$$
We introduced the arbitrary phase difference to be more general. Mixing this signals we get
$$s(t) = x_1(t)x_2(t) = e^{j{\pi}\frac{\beta}{\tau}t^2}e^{j({\pi}\frac{\beta}{\tau}t^2 + \phi)} = e^{j({\pi}\frac{2\beta}{\tau}t^2 + \phi)}$$
We see that the chirp rate is now $2\beta/\tau$, so the resulting chirp will will change frequency twice as fast, and have a phase offset of $\phi$.
FMCW Radar
Resolving the return using a FMCW radar is a different story. In order to simulate it, you must introduce the target's delay. The mixing process is very similar, but with a slight tweak.
Let the chirped signal we transmit be
$$s_{tx}(t) = e^{j\pi\frac{\beta}{\tau}t^2}$$
After reflecting from a target we receive the signal after some delay $t_d$, we have
$$s_{rx}(t) = e^{j\pi\frac{\beta}{\tau}(t - t_d)^2} = e^{j\pi\frac{\beta}{\tau}(t^2 - 2tt_d + t_d^2)}$$
After mixing $s_{rx}(t)$ with the complex conjugate of $s_{tx}(t)$, the higher order term containing $e^{j\pi\frac{\beta}{\tau}t^2}$ drops off and we're left with
$$x(t) = e^{j\pi\frac{\beta}{\tau}(-2tt_d + t_d^2)} = e^{-j\pi\frac{\beta}{\tau}2tt_d}\,e^{j\pi\frac{\beta}{\tau}t_d^2}$$
Equating terms this result is equivalent to
$$e^{j({2\pi}f_bt+ \phi)}$$
This is simply a sinusoid at the beat frequency $f_b$. We can ignore the $\phi$ term since it will have no effect on our ability to resolve the target's range. Equating the the phase functions
$$-\pi\frac{\beta}{\tau}2tt_d = 2{\pi}f_bt$$
So that then we have
$$f_b = -\frac{\beta}{\tau}t_d$$
Since we know our pulse travels at the speed of light $c$, we can rewrite the target's delay in terms of range $R$ and yield the mapping between target range and its beat frequency
$$t_d = \frac{2R}{c} => f_b = -\frac{2R\beta}{c\tau}$$
Update
I had promised to augment to show how this works using real signals only, which brings up the need to filter out terms post-mixing before doing the Fourier transform on the dechirped signal. Better late than never.
Let's take the same parameters of the chirp, except now the transmitted and receive signals are
$$s_{tx}(t) = cos(\pi\frac{\beta}{\tau}t^2)$$
$$s_{rx}(t) = cos(\pi\frac{\beta}{\tau}(t - t_d)^2)$$
When multiplying sinusoids to perform dechirping, in this case a cosine, the product property is
$$cos(\alpha)cos(\beta) = \frac{1}{2}(cos(\alpha - \beta) + cos(\alpha + \beta))$$
Using $s_{tx}(t)$ and $s_{rx}(t)$ above we then get
$$s_{rx}(t)s_{tx}(t) = cos(\pi\frac{\beta}{\tau}(2t^2 - 2tt_d + t_d^2)) + cos(\pi\frac{\beta}{\tau}(-2tt_d + t_d^2))$$
Due to the mixing process producing both a sum and difference arguments in the cosine terms, we only want the differnce, as the first term will contain unwanted frequency information. For this, we use a low-pass filter to only let the second term though. The final signal is then
$$x(t) = cos(\pi\frac{\beta}{\tau}(-2tt_d + t_d^2))$$
This is the real part of the $x(t)$ described in the original post. Taking the Fourier transform now will give you the beat frequencies as desired, albeit you will have both negative and positive frequency components at $f_b = \pm\frac{2R\beta}{c\tau}$. You can discard half of the spectrum and continue as usual.
