Why is enthalpy and not heat released by system used while calculating entropy of surroundings?
Because $q=\Delta H$ only if only expansion work is done and the pressure of the system is constant. In that case, for an isothermal process, in which T is the same for system and surroundings, the entropy of the surroundings can be equated with the change in enthalpy divided by T. Which is why the applicability of the integrated formula $\Delta G = \Delta H -T \Delta S$ is restricted to constant p and T. You could write $q_p$ in place of $\Delta H$ the equation but only when work is restricted to pV work and when T and p are constant. It would not hold generally since sometimes work other than expansion work is performed. In general $\Delta H = q_p+w_\textrm{non-pV}$ at constant T and p.
To show this more explicitly, for a small (differential) change:
$$dU=dq+dw$$
$$dU=dq+dw_{\textrm{non-pV}}+dw_{\textrm{pV}}$$
$$dH = dU+d(pV) = dq+dw_{\textrm{non-pV}}+dw_{\textrm{pV}} + pdV+Vdp$$
At constant pressure of the system $dw_{\textrm{pV}}=-pdV$, $Vdp=0$ and
$$dH = dq+dw_{\textrm{non-pV}}$$
Further derivation is possible* to show how this relates to $G$, by assuming constant T and considering reversibility, but this should suffice to show that $dH$ and $dq$ (or $\Delta H$ and $q$) are not in general the same.

*For a reversible process $dq=TdS$ and
$$dH =  TdS+dw_{\textrm{non-pV,rev}}$$
Finally the differential expression for $G$ at constant $T$ follows (since then $dG=dH-d(TS)=dH-TdS$):
$$dG=dH - TdS = dw_{\textrm{non-pV,rev}} $$