How does a change in base current affect collector current once the transistor is biased
In what way do I have to look at this to be able to understand?


The base current increase (decrease) is due to an increase (decrease) in \$v_{BE}\$.

The increase (decrease) in \$v_{BE}\$ increases (decreases) the injection of carriers from the heavily doped emitter.

Most of these carriers cross the thin base region without recombining and are then swept across the base-collector junction into the collector region.  A small percentage don't and these form the base current.

Update to answer a comment:


  Once it has been biased isnt Vbe=0.7V. Now when we apply a small ac
  signal of the order of milli volt isnt the change negligible?


No, the collector current is exponential in the base-emitter voltage:

\$i_C = I_Se^{(v_{BE}/V_T)}\$

To get a feel for this, consider this question:  to double the collector current, how much would \$v_{BE}\$ need to increase?

For example, assuming the bias value is \$V_{BE} = 0.7V \$, increasing this voltage by a mere \$17 mV \$ (an increase of just under 2.5%) will double the collector current.

Another approach:

According to the collector current equation, if we change the base-emitter voltage from its quiescent value by some small amount, the change in collector current is approximately:

\$\Delta i_C = \dfrac{I_C}{V_T} \Delta v_{BE}\$

As a typical example, let the quiescent collector current \$I_C = 1mA \$.  At room temperature, \$V_T = 25mV\$.

Then, for these numbers, the collector current changes by 4% when the base-emitter voltage changes by 1mV.