I'm afraid I am not sure what exactly it is you don't understand. Perhaps an example that's easy enough to calculate by hand will illustrate.
Suppose you have the following raw data:
5,4 are the raw values making up SET1. Thus N1 = 2, M1 = 4.5 and SD1 = 0.7071.
3,2,4,7 are the raw values making up SET2. Thus N2 = 4, M2 = 4 and SD2 = 2.1602.
Now calculate the D-score nominator as well as Cohen's pooled SD:
D_asis_denom = 1.7248
Cohen_pooled = 1.9039
Also calculate the SD across all six raw data points, i.e. SDALL = sd(5,4,3,2,4,7). The result is
SDALL = 1.7224.
You'll notice that SDALL is -- disregarding rounding errors, etc. -- identical to D_asis_denom, but noticeably different from the pooled SD for Cohen's d.
In SPSS syntax:
COMPUTE N1 = 2.
COMPUTE M1 = mean(5,4).
COMPUTE N2 = 4.
COMPUTE M2 = mean(3,2,4,7).
COMPUTE D_asis_denom = SQRT( ( ((N1-1) * SD1**2 + (N2-1) * SD2**2) + ((N1+N2) * ((M1-M2)**2) / 4) ) / (N1 + N2 - 1) ).
COMPUTE Cohen_pooled = SQRT( ((N1-1) * SD1**2 + (N2-1) * SD2**2) / (N1 + N2 - 2) ) .