2. A.

Notice that,
$$\begin{aligned} t & =2 x y+y z+2 z x=2 x(y+z)+y z \\ & \leqslant 2 x(y+z)+\frac{1}{4}(y+z)^{2} \\ & =2 x(1-x)+\frac{1}{4}(1-x)^{2} \\ & =-\frac{7}{4}\left(x-\frac{3}{7}\right)^{2}+\frac{4}{7} . \end{aligned}$$

When $x=\frac{3}{7}, y=z=\frac{2}{7}$, the equality holds.
At this point, $t=2 x y+y z+2 z x$ achieves its maximum value $\frac{4}{7}$.