Lithium Battery Charger using LM317
Does R3 affect the LM317 output voltage as it's in series with R2?


If the battery is not connected (or, equivalently, if there is only a negligible current through the battery), the circuit operates as a constant voltage source with an output voltage of \$V_{0}=V_{ref}\cdot \left( 1 + \frac{R_2+R_3}{R_1}\right) + I_{Adj}\cdot (R_2+R_3)\$. Since \$R_3\$ is typically much smaller than \$R_2\$, \$R_3\$ has only a negligible influence on \$V_0\$. On the other hand, if a battery is connected, as long as the battery voltage is lower than \$V_0\$, the circuit works as a constant current source and the current through the battery will be approximately \$\frac{0.6V}{R_3}\$. In this case, the voltage across the battery will automatically be adjusted such that the current remains constant. Once the battery voltage reaches \$V_0\$, the current can no longer be kept constant. In this case, the voltage across the battery will be constant, namely \$V_0\$, independent on the current that flows through the battery. 


  How does the BJT and resistor precisely work in this setup? Won't the battery initially try and draw as much current as it can, meaning 0.6v would develop across R3 before the battery is sufficiently charged to transition into the CV stage?


Yes, the battery will try to draw as much current as it can, but it will not succeed, because the higher the current through the battery, the higher the voltage across \$R_3\$. When this voltage reaches 0.6V, the transistor starts to conduct and the output voltage of the regulator will decrease. Therefore, the current through the battery is automatically limited to approximately \$\frac{0.6V}{R_3}\$. 


  Wouldn't the 0.6V developed across R3 mean that the potential "seen" by the battery is approx 6.3V? (6.9V LM317 output subtract 0.6V). Wouldn't this be insufficient to charge the battery fully?


Note that the battery is connected between the output of the voltage regulator and the base of the transistor, thus the voltage across \$R_3\$ has no influence on the battery voltage. The voltage regulator makes sure that the voltage across the battery will not exceed \$V_0\$. 


  When the BJT turns on, the voltage output of the LM317 will drop to 1.25V as the adjust pin is shorted (which bypasses the resistors). Surely this is not suitable for the CV stage of charging as the potential drops to way below the battery charging voltage? How is current output to the battery affected during this voltage drop?


The transistor will not conduct completely. The control loop ensures that the current through the transistor is just large enough to maintain a constant current. During charging with a constant current, the voltage of the battery will slowly increase. Once it reaches \$V_{0}\$ the current can no longer be kept constant and the voltage will remain at \$V_{0}\$.

However When charging lithium batteries, switching from CC to CV must take place at a precisely defined voltage. Therefore, \$R_2\$ should be made adjustable in order to set \$V_0\$ to the required level. This circuit is interesting to analyze and it might be a cheap solution, but there a integrated circuits that do a better job. 

The simulation (using a rather crude battery model) shows the relationship between battery voltage (green) and current through the battery (blue). The constant current source does not show ideal behaviour.