It is better to "parse" these networks from the output back towards the input, calling their input some general $x$ and performing substitutions and/or compositions.
So, let's call these networks $U$pper and $L$ower.
From the upper diagram:
$$UH_2[n] = x[n] + x[n-1] \cdot -a_1$$
$$UH_1[n] = x[n] \cdot b_0+x[n-1] \cdot b_1$$
Now, the output of $U$pper is given by the composition of the two:
$$UH_y[n] = UH_2[UH_1[n]]$$
This is because, the output of one, becomes the input to the other.
$$UH_y[n] = UH_1[n] + UH_1[n-1] \cdot -a_1$$
...and if you substitute...
$$UH_y[n] = x[n] \cdot b_0 + x[n-1] \cdot b_1 + (x[n-1] \cdot b_0 + x[n-2] \cdot b_1) \cdot -a_1 \Rightarrow \\ x[n] \cdot b_0 + x[n-1] \cdot b_1 + x[n-1] \cdot b_0 \cdot -a_1 + x[n-2] \cdot b_1 \cdot -a_1$$
Notice here that if you are trying to get to the $-1$ of the $x[n-1]$, then that would be the $x[n-2]$.
And this concludes the $U$pper part.
For the $L$ower part, we are not going to go through the whole thing, because of two reasons:
If you notice, the $H2, H1$ are reversed (The $L$ower network calls $H2$ what the $U$pper network calls $H1$).
We have already mapped $H2, H1$.
So, the lower network's response is:
$$LH_y[n] = LH_1[LH_2[n]]$$
$$LH_y[n] = LH_2[n] \cdot b_0+LH_2[n-1] \cdot b_1$$
And if you substitute:
$$LH_y[n] = (x[n]+ -a_1 \cdot x[n-1]) \cdot b_0 + (x[n-1]+ -a_1 \cdot x[n-2]) \cdot b_1 \Rightarrow \\
x[n] \cdot b_0 + -a_1 \cdot b_0 \cdot x[n-1] + x[n-1] \cdot b_1 + b_1 \cdot -a_1 \cdot x[n-2]$$
Which, after you re-arrange, looks exactly the same.
Hope this helps.