Are the coefficients and exponents of contracted gaussian same for cartesian and pure form?
They are the same.

Your equation can be rewritten as

$$ \phi_\mathbf l^\text{CGF}(r,\Omega)=\frac{x^{l_x}y^{l_y}z^{l_z}}{r^l}\sum_ns_nr^l\exp(-\alpha_nr^2)\equiv\Phi_\mathbf l(\Omega)R_l(r) $$

where $l=l_x+l_y+l_z$ is the total angular momentum, $\Phi_\mathbf l(\Omega)$ is the angular part, and $R_l(r)$ is the radial part, and they are separable not only in mathematical form, but also for practical purposes. (This discussion omits the norm, which is not separable to radial and angular contributions.) When specifying a Gaussian basis set in quantum chemistry, one talks only about the radial part, independent of the angular part. The angular part is really considered an implementation detail, because its form usually doesn't have any significant effect on the results.

For completeness, two most common angular forms are Cartesian:

$$ \Phi_\mathbf l(\Omega)=\frac{x^{l_x}y^{l_y}z^{l_z}}{r^l} $$

and spherical harmonic:

$$ \Phi_\mathbf l(\Omega)=Y_l^m(\theta,\varphi) $$