$\Delta_r \mathrm{H}$ and $\Delta_r \mathrm{S}$ are independent of each other
Processes can be exothermic or endothermic, and they can show a decrease or increase in entropy. You can't predict $\Delta_r \mathrm{H}$ from $\Delta_r \mathrm{S}$ or vice versa. You can combine the two to calculated the change in Gibbs energy, and this tells you about the direction the process will go in the absence of non-PV work.
$\Delta_r \mathrm{H}$ and $\Delta \mathrm{S}_\mathrm{surr}$ are linked under certain circumstances
In the absence of non-PV work, all energy exchange between reaction and surrounding will be through PV work and through heat exchange. With these restrictions and when there are no differences in the temperature before and after the reaction, $\Delta_r \mathrm{H}$ will be equal to the heat transfer to the surroundings. If the surroundings are a giant reservoir, its temperature will not change much (for the ideal case, not at all), and the heat exchange will almost be reversible (for the ideal case, reversible). In that case,
$$\Delta \mathrm{S}_\mathrm{surr} = - \frac{\Delta_r \mathrm{H}}{T}$$
$\Delta_r \mathrm{S}$ and $\Delta \mathrm{S}_\mathrm{surr}$ are independent of each other
Because $\Delta_r \mathrm{S}$ and $\Delta_r \mathrm{H}$ are independent of each other, $\Delta_r \mathrm{S}$ and $\Delta \mathrm{S}_\mathrm{surr}$ are, too. The one thing that connects them is that processes for which the sum of the entropy changes is smaller than zero don't happen because that would go against the second law of thermodynamics.
Thus, shouldn't the total entropy change be zero?
No, it shouldn't. If it is, you have reached equilibrium and there would not be any changes in any state functions at all. The reaction in question, however, is described as spontaneous. When a spontaneous reaction proceeds, the total entropy will increase.