I will sketch just the idea omitting several (actually important) mathematical details.

If $F=F[\psi]$ is a functional, we define, if exists, its functional derivative $\frac{\delta F}{\delta \psi}$ as the function (a distribution, more generally) such that

$$\left.\frac{dF[\psi + h \phi]}{dh}\right|_{h=0}= \int \frac{\delta F}{\delta \psi}(x) \phi(x) dx$$

for every test function $\phi$. Now suppose that $\psi$ parametrically depends on the parameter $\theta$. We have

$$\frac{dF[\psi_\theta]}{d\theta} =\lim_{h\to 0} \frac{1}{h}\left(F[\psi_{\theta + h}] - F[\psi_\theta]\right)$$

$$=\lim_{h\to 0} \frac{1}{h}\left(F[\psi_\theta + h \partial_\theta \psi + O_\theta(h^2)] - F[\psi_\theta]\right)$$

$$= \lim_{h\to 0} \frac{1}{h}\left(F[\psi_\theta + h \partial_\theta \psi] - F[\psi_\theta]\right) \:.$$

$$= \left.\frac{dF[\psi_\theta + h \partial_\theta \psi_\theta]}{dh}\right|_{h=0}\:.$$

Applying the definition of functional derivative given above, we have that

$$\frac{dF[\psi_\theta]}{d\theta} = \left.\frac{dF[\psi_\theta + h \partial_\theta \psi_\theta]}{dh}\right|_{h=0}= \int \frac{\delta F}{\delta \psi_\theta}(x) \partial_\theta\psi_\theta (x) dx\:.$$

There are several open mathematical issues in the outlined procedure (for instance dropping $O_\theta(h^2)$ in the formula above is not so easy as it could seem at first glance). Nevertheless, everything goes right (it can be proved by direct inspection) when dealing with functionals $F[\psi]$ of integral form like the one you consider and assuming to work with a domain of suitably smooth and rapidly vanishing at infinity functions.