Transformed Direct Form II filter: Z transform: reduce multiplications
It's instructive to look at the actual difference equation implemented by the system. If you define a sequence $w[n]$ as the output of the first (left-hand side) adder you get

$$w[n] = x[n] + Aw[n-1]\\
y[n] = Bw[n] + Bw[n-1]$$

Applying the $\mathcal{Z}$-transform gives

$$W(z) = X(z) + A \ W(z)z^{-1}\\
Y(z) = B \ W(z)\left(1 + z^{-1}\right)$$

which results in

$$W(z) = \frac{X(z)}{1-Az^{-1}}\\
Y(z) = \frac{B(1+z^{-1})}{1-Az^{-1}}X(z)$$

In the time-domain this last equation is equivalent to

$$y[n] = Ay[n-1]+B \left( x[n] + x[n-1] \right) \tag{1}$$

Equation (1) shows that this system can obviously be implemented with only two multiplications. If you want to use the Transposed Direct-Form II structure then just use the structure shown here (bottom figure), set $a_1=-A$, remove the branches with multipliers $a_2$ and $b_2$ and the lower delay element (because you only have a first order filter), remove the multipliers $b_0$ and $b_1$, and multiply the input signal directly with $B$.

EDIT: (in reaction to your updated question)

There are a few mistakes in your figure. Note that in the figure I linked to, the input is on the right-hand side. This might have caused some confusion. Anyway, in your figure where you multiply by $B$, just multiply by $A$ instead. Then remove the multiplication by $-A$ and add a multiplication by $B$ to the input signal $x[n]$, before the path splits. The node you refer to is simply an addition; they obviously forgot the 'plus' sign.

As for the equations, what I wanted to do is to get a difference equation for the output signal $y[n]$ in terms of $x[n]$ and of delayed versions of $x[n]$ and $y[n]$. This shows what the system actually does and in this way the description of the system becomes independent of the chosen filter structure. In order to do that, I used the $\mathcal{Z}$-transform which transforms difference equations into algebraic equations. So in the $\mathcal{Z}$-domain it is straightforward to eliminate the signal $w[n]$ and write $Y(z)$ directly in terms of $X(z)$. Transforming back to the time-domain yields the desired difference equation, which can then be implemented by any structure you like.