The Berezin path integral can be viewed as an infinite product of Berezin integrals. The notation is the same as for regular integrals against a measure because it just works like it (modulo a subtlety for changes of variables) when it comes to defining a kind of integration measure. So let us first think of a complex scalar field:
\begin{equation}
\int \mathcal{D}[\mathrm{Re}(\phi)]\mathcal{D}[\mathrm{Im}(\phi)]f[\phi] \equiv \left(\int \prod_{x\in \mathbb{R}^{1,3}}d[\mathrm{Re}(\phi_x)]d[\mathrm{Im}(\phi_x)]\right)f[\phi]. \tag{1}
\end{equation}
Where the subscript means we evaluate the field at $x$ (to avoid nested brackets for clarity). Now, we can use a change of variable:
\begin{align}
&\mathrm{Re}(\phi_x) \leadsto \mathrm{Re}(\phi_x)+i\mathrm{Im}(\phi_x) = \phi_x, \tag{2}
\\
&\mathrm{Im}(\phi_x) \leadsto -i\mathrm{Im}(\phi_x)+\mathrm{Re}(\phi_x) = \overline{\phi}_x. \tag{3}
\end{align}
The determinant of this transformation is a constant that we now absorb in the integration measure $d[\mathrm{Re}(\phi_x)]d[\mathrm{Im}(\phi_x)] \leadsto d\phi_x d\overline{\phi}_x$. The purpose of not showing the constant is that, since $\mathcal{D}\phi$ is a purely notational convention, it can absorb any constant we want, as long as we don't integrate over a possible dependence on another field (like in the case where one would eventually also integrate over a gauge field, as your notation could suggest). Anyway, we can write:
\begin{equation}
\left(\int \prod_{x\in \mathbb{R}^{1,3}}d[\mathrm{Re}(\phi_x)]d[\mathrm{Im}(\phi_x)]\right)f[\phi] \leadsto \left(\int \prod_{x\in \mathbb{R}^{1,3}}d\phi_x d\overline{\phi}_x\right)f[\phi] \equiv \int \mathcal{D}\phi \mathcal{D}\overline{\phi}f[\phi]. \tag{4}
\end{equation}
For Berezin integrals, this is formally the same procedure except for the determinant, which is in the denominator now! But since we absorb it in the integration measure, we won't see any difference. Let $\psi$ be a Dirac bi-spinor, and $\psi^\dagger$ its transpose-conjugate. We have:
\begin{equation}
\psi\equiv \left(\begin{matrix} \theta_1\\ \theta_2 \\ \chi_1 \\ \chi_2 \end{matrix} \right),\,\,\,\psi^\dagger \equiv \left(\begin{matrix} \overline{\theta}_1 & \overline{\theta}_2 & \overline{\chi}_1 & \overline{\chi}_2 \end{matrix} \right). \tag{5}
\end{equation}
With $(\theta_1,\theta_2)$ and $(\chi_1,\chi_2)$ two pairs of complex Grassmann-odd-valued fields (sometimes, one writes instead $(-\chi_2,\chi_1)$ to emphasize the $\mathbb{C}$-structure of the space of bi-spinors.) Then, defining the Berezin path integral as:
\begin{equation}
\left(\int \prod_{i=1}^2\prod_{x\in \mathbb{R}^{1,3}}d[\mathrm{Re}(\theta_{i,x})]d[\mathrm{Im}(\theta_{i,x})]\right)\left(\int \prod_{i=1}^2\prod_{x\in \mathbb{R}^{1,3}}d[\mathrm{Re}(\chi_{i,x})]d[\mathrm{Im}(\chi_{i,x})]\right)f[\psi]. \tag{6}
\end{equation}
Analogously to the previous case of a real scalar field, we can bring this expression into a form:
\begin{equation}
(6) \leadsto \left(\int \prod_{i=1}^2\prod_{x\in \mathbb{R}^{1,3}}d\theta_{i,x}d \overline{\theta}_{i,x}\right)\left(\int \prod_{i=1}^2\prod_{x\in \mathbb{R}^{1,3}}d\chi_{i,x}d\overline{\chi}_{i,x}\right)f[\psi] \equiv \int \mathcal{D}\psi \mathcal{D}\psi^\dagger f[\psi]. \tag{7}
\end{equation}
Now, for the Dirac conjugate instead of the transpose-complex one, just use the determinant of $\gamma^0$ (in the denominator because we work with anti-commuting numbers), which is $1$. So finally we have defined (at least at the physicist's level) something of the form:
\begin{equation}
\int \mathcal{D} \psi \mathcal{D} \overline{\psi}\,f[\psi]. \tag{8}
\end{equation}
To conclude, I should say that technically, writing $\mathcal{D} \psi \mathcal{D} \overline{\psi}$ or $\mathcal{D} \overline{\psi} \mathcal{D} \psi$ is a matter of convention. The idea behind not using the real and imaginary parts is 1) for notational convenience and 2) because we can treat the fields and their conjugate as independent variables. So the notation $\mathcal{D} \psi \mathcal{D} \overline{\psi}$ makes sense.
