Stage 1 is a MUX - finding the output should be straightforward

$Y = \overline{S_1}.\overline{S_0}.1 + \overline{S_1}.{S_0}.0 + {S_1}.\overline{S_0}.0 + {S_1}.{S_0}.1$

substitute for $S_1, S_0$

$Y = \overline{A}\overline{C} + {A}{C}$

Move on to the decoder now. We're only bothered with the outputs $O_0$ and $O_2$, lets not bother with any others:

$O_0 = \overline{B}\overline{Y}$, think about why this is. The decoder output $O_0$ is selected when input is $00$ (converts encoded binary to decoded decimal)

$O_2 = \overline{B}{Y}$, Output $O_2$ selected when input is $10$

And the outputs are bubbled (negated) before fanning into the NAND:

$f(A,B,C) = \overline{\overline{O_0}.\overline{O_2}} = O_0 + O_2$

Now just substitute for Y in the expressions for $O_0 + O_2$ we obtained earlier, and you will get the required minterms of $\Sigma(0, 1, 4, 5)$