How does the voltage divider model a voltage source in this transistor bias circuit?
A voltage divider produces a voltage at the junction between the two resistors and the value of that voltage is easy to calculate, given the source voltage and resistor values. Now, if you connect something else to the junction between the two resistors (such as the base of your BJT) then you need to worry about how much current will be drawn by the thing you have connected. As long as the current drawn by the transistor's base is much less than the current that would flow through the resistors without the base connection, then the resistor divider is a reasonable approximation of an ideal voltage source.

So, in this case you just need to select resistor values for the voltage divider so that


the resulting voltage is equivalent to the battery in the left-hand
circuit, and
the current through the resistors is much greater than the base current.