7. If $x, y, z > 0$ satisfy
$$
\left\{\begin{array}{l}
\frac{2}{5} \leqslant z \leqslant \min \{x, y\}, \\
x z \geqslant \frac{4}{15}, \\
y z \geqslant \frac{1}{5},
\end{array}\right.
$$
7.13.

From the problem, we have
$$
\frac{1}{\sqrt{x}} \leqslant \frac{\sqrt{15 z}}{2}, \frac{1}{z} \leqslant \frac{5}{2}, \frac{1}{\sqrt{y}} \leqslant \sqrt{5 z} \text {. }
$$

Then 
$$
f=\frac{2}{\sqrt{x}} \cdot \frac{1}{\sqrt{x}}+\frac{1}{z}\left(1-\frac{z}{x}\right)+
$$
$$
\begin{aligned}
& 2\left[\frac{2}{\sqrt{y}} \cdot \frac{1}{\sqrt{y}}+\frac{1}{z}\left(1-\frac{z}{y}\right)\right] \\
\leqslant & \frac{2}{\sqrt{x}} \cdot \frac{\sqrt{15 z}}{2}+\frac{5}{2}\left(1-\frac{z}{x}\right)+ \\
& 2\left[\frac{2}{\sqrt{y}} \cdot \sqrt{5 z}+\frac{5}{2}\left(1-\frac{z}{y}\right)\right] \\
= & \frac{5}{2}+\sqrt{15} \cdot \sqrt{\frac{z}{x}}-\frac{5}{2} \cdot \frac{z}{x}+ \\
& 2\left[\frac{5}{2}+2 \sqrt{5} \cdot \sqrt{\frac{z}{y}}-\frac{5}{2} \cdot \frac{z}{y}\right] \\
= & 13-\frac{5}{2}\left(\sqrt{\frac{z}{x}}-\sqrt{\frac{3}{5}}\right)^{2}-5\left(\sqrt{\frac{z}{y}}-\frac{2}{\sqrt{5}}\right)^{2}
\end{aligned}
$$
$$
\leqslant 13 \text {. }
$$

When $(x, y, z)=\left(\frac{2}{3}, \frac{1}{2}, \frac{2}{5}\right)$, the equality holds. Therefore, $f_{\max }=13$.