How does full wheatstone bridge with strain gauges work?
1: It will always be directly proportional to the weight applied.

A strain gauge has a linear response, and elastic bending of a load cell is linear.  Thus, we can write \$ r_i = k_i F \$ where \$k_i\$ is a constant of proportionality for that gauge, and \$F$ the force applied.

If we substitute and rearrange, we can take out \$F\$ and the last line of your question becomes:

\$=E F \left\{ \frac{(k_2-k_1 -k_4 + k_3)}{4R} \right\} \$

i.e. it is directly proportional to \$F\$. This still assumes \$r_i \ll R \forall i\$ of course.

2: You choose where to put the strain gauges

The constant of proportionality above has a \$(k_2-k_1 -k_4 + k_3)\$ term. If all of these \$k_i\$ are the same, then it sums to zero. That isn't any use. To make the constant of proportionality large, we want \$k_2\$ and \$k_3\$ to be as large as possible, and \$k_1\$ and \$k_4\$ to be as negative as possible. Strain gauges with negative responses are hard to find, but the metal load cell will have some areas under tension, and some under compression. So we usually choose to mount two strain gauges in an area under compression, and the other two in an area under tension.