Intuitive is a matter of what works well in one's own mind. In this case, I can only illustrate one method that comes to my mind. (This isn't the only approach, though.) It's simply an algebraic one:
$$\begin{align*}
y &=(x_1' + x_3\, x_4\, x_5)\:(x_1'\, x_2' + x_3\, x_4)
\\
&=((x_1' + x_3\, x_4\, x_5)\:(x_1'\, x_2' + x_3\, x_4))''\tag{1}
\\
&=((x_1' + x_3\, x_4\, x_5)'+(x_1'\, x_2' + x_3\, x_4)')'\tag{2}
\\
&=(x_1 \,(x_3'+ x_4'+ x_5')+(x_1+ x_2)\,(x_3'+ x_4'))'\tag{3}
\\
&=(x_1 \,x_3'+ x_1 \,x_4'+ x_1 \,x_5'+ x_1\,x_3'+ x_1\,x_4' + x_2\,x_3'+ x_2\,x_4')'\tag{4}
\\
&=(x_1 \,x_3'+ x_1 \,x_4'+ x_1 \,x_5' + x_2\,x_3'+ x_2\,x_4')'\tag{5}
\\
&=(x_1 \,(x_3'+ x_4'+ x_5') + x_2\,(x_3'+ x_4'))'\tag{6}
\\\\
\text{Set } t_1 &=x_3'+ x_4'= (x_3 \, x_4)'
\\\\
&= (x_1 \,(t_1+ x_5') + x_2\,t_1)'\tag{7}
\\\\
\text{Set } t_2 &=t_1 + x_5'= (t_1' \, x_5)'
\\\\
&= (x_1 \,t_2 + x_2\,t_1)'\tag{8}
\\
&= ((x_1 \,t_2)' \,(x_2\,t_1)')''\tag{9}
\end{align*}$$
From here you can see that you'll need two NOTs and five NANDs.