The main point to takeaway is the relation \$ I_C = \beta I_B\$ is a model of how a BJT behaves when operating in forward-active mode. As an extreme example, one cannot simply apply a negative current to the base of an NPN BJT and expect negative \$\beta I_B\$ current to flow in the collector.
The actual reason current flows in a BJT is extremely complicated and only described by modern solid state physics. However the models that have been developed to describe their behavior are in most cases "good enough" when applied within their regions of validity.
I have simplified your schematic slightly to illustrate the concept of saturation.
Take for example the following schematic,
Here the current into the base terminal is set by an ideal current source (only for convince in a Spice simulator) in a real circuit a resistor is often close enough to set the base current by ohms law.
The following figure is the simulations reults of a DC sweep of the current source injecting into the base terminal. The x-axis is base-current. The left y-axis is the collector voltage. The right y-axis is the collector-current.
We can see that when the collector-emitter voltage is greater than ~0.2V the forward-active relation holds true, namely, \$I_C = \beta I_B\$. As more base-current is applied more collector-current flows. However as more current flows more voltage is being dropped across R1 and the collector-emitter voltage is dropping. Eventually when the collector-emitter voltage becomes small enough the BJT is no longer in forward-active but in forward-saturation.
The textbook definition of forward-saturation for an NPN BJT is when Vbe is positive and when Vbc is also positive.
In saturation the BJT, loosely speaking it looks like a switch between collector and emitter. Slightly more complete model is a small mV level voltage source with low series resistance.
So when the NPN is behaving as a switch (saturated) it is the circuit around it that dictates what current will flow, in your case R1.