Sub-band factor 0.5 in uniform filterbank
The factor of 1/2 comes from the fact that you are down-sampling. The down-sampling property of the z-transform states that given:

$$ y[n] = x[nL] $$

Then:

$$ Y(z) = \frac{1}L \sum_{r=0}^{L-1}X(z^{1/L}e^{-j2{\pi}r/L})$$

And in your case $L = 2$ so $r$ indexes from 0 to 1 so then:

$$ Y(z) = \frac{1}2 \sum_{r=0}^{1}X(z^{1/2}e^{-j{\pi}r})$$

To be more thorough:

$$ y[n] = u_i[n] = (x * h_i)[nL] $$

So:

$$ Y(z) = U_i(z) = \frac{1}L \sum_{r=0}^{L-1}H_i(z^{1/L}e^{-j2{\pi}r/L})X(z^{1/L}e^{-j2{\pi}r/L})$$

And again $L = 2$ so $r$ indexes from 0 to 1 so then:

$$ Y(z) = U_i(z) = \frac{1}2 \sum_{r=0}^{1}H_i(z^{1/2}e^{-j{\pi}r})X(z^{1/2}e^{-j{\pi}r})$$