One of the purposes of defining a system and its surrounding in thermodynamics is to separate the interesting stuff from the boring stuff.
$\Delta S_\text{surroundings}$ can be re-expressed as $\dfrac{-\Delta H_\text{system}}{T}$
The best boring surrounding is a large homogeneous phase with the same temperature as the system, for example a stirred water bath. (If the temperature of the system changes, you have to quickly swap the bath to match the temperature.) In this specific scenario, the entropy change of the surrounding will be equal to the heat $-q$ transferred into the surrounding divided by the temperature. If the process in the system is happening at constant pressure, the system is closed and there is no non-PV work (no electrical, mechanical, etc. work, just expansion work), then the heat $q$ transferred out of the system will be equal to the change in enthalpy of the system, $\Delta H$.
So there are a lot of disclaimers to make before you can set $\Delta S_\text{surroundings} = \dfrac{-\Delta H_\text{system}}{T}$.
[...] why $\Delta S_\text{system}$ cannot also be re-expressed (or at least is never written) as a term with some combination of $\Delta H$ and $T$
Remember we want the interesting stuff to happen inside the system. In general, there is no way of predicting the entropy change in the system from other thermodynamic values unless the system is as boring as the surrounding (for example when the system is a block of hot metal and the surrounding a block of cold metal). Of course, if you put the interesting stuff in the surrounding and keep the system boring, $$\Delta S_\text{system} = \dfrac{-\Delta H_\text{surroundings}}{T}$$ would be appropriate.
Is ΔS of a system related to temperature and change in enthalpy?
In general, it is not. Entropy and enthalpy change are not coupled. They can both be negative, both positive, one negative and the other positive and vice versa. This shows you that there is no fixed relationship between entropy and enthalpy.
There is one special case where entropy change and enthalpy change of the system are related - when the change in Gibbs energy is zero. The change in Gibbs energy is expressed in terms of entropy and enthalpy changes as (in the following, the subscript "system" is omitted because all the quantities refer to the system):
$$\Delta G = \Delta H - T \Delta S$$
When $\Delta G$ is zero (for example when a substance is at its melting temperature and both solid and liquid phase are present), the following relationship holds:
$$ \Delta S = \frac{\Delta H}{T}$$
Notice that there is no negative sign. This relationship can be used to estimate the melting point of a substance. Notice also that for chemical reactions at equilibrium, $\Delta_r G^\circ$ is not equal to zero, but $\Delta_r G$ is.
So while at equilibrium,
$$\Delta_r S = \Delta_r S^\circ + R \ln K = \frac{\Delta_r H}{T}$$
is correct, you can't calculate $\Delta_r S^\circ$ from $\Delta_r H^\circ$, i.e.
$$ \Delta_r S^\circ \neq \frac{\Delta_r H^\circ}{T}.$$
The only exception is when $K$ happens to be one, as it is by definition for melting when the substance is at the melting point.
$ΔS_\text{system}$ is usually given as a standard value, seemingly independent of temperature
Both $ΔS_\text{system}$ and $ΔH_\text{system}$ are dependent on temperature, and the heat capacity is used to describe the temperature-dependency. If you are evaluating the temperature-dependency of $ΔG_\text{system}$, however, the largest dependency comes from having the temperature directly in the defining equation:
$$\Delta G = \Delta H - T \Delta S$$
So even if entropy and enthalpy were temperature-independent (again, they are not), $\Delta G$ would still depend significantly on temperature. Oddly enough, a rigorous treatment of the temperature-dependence of $\Delta G$ gives the result that you would get from differentiating the defining equation with respect to $T$:
$$\frac{d \Delta G}{dT} = - \Delta S$$