Why May K Expressions Be Manipulated As Such? Why is the above massaging of various K expressions valid? Something in me says "Hess's Law." But then doesn't that deal with enthalpy? What's the true reason?
Your intuition is correct. The reason equilibrium constants from different reactions can be strung together and multiplied comes from a generalization of Hess' law which covers Gibbs' free energy changes (apparently also called Bordwell thermodynamic cycles). Enthalpy is a thermodynamic function with a special property; it is a state function. That is, enthalpy changes in cycles are determined solely by the initial and end states of a system, and not the trajectory in which the initial state travels to the end state (as happens for functions such as work $W$ and heat $q$). Any state function can be manipulated in the way enthalpy is used for Hess diagrams. Thus, by adding reactions, you can add their respective reaction free energy changes $\Delta_rG$.

Let us now take two reactions, with free energy changes equal to $\Delta_rG_1$ and $\Delta_rG_2$. There is a relationship between a reaction's free energy change and its reaction quotient $Q$, namely:

$$\Delta_rG = \Delta_rG^o + RT\ ln\ Q$$

, where $^o$ indicates standard conditions. For reactions 1 and 2, we therefore have:

$$\Delta_rG_1 = \Delta_rG^o_1 + RT\ ln\ Q_1$$
$$\Delta_rG_2 = \Delta_rG^o_2 + RT\ ln\ Q_2$$

Adding the reactions, we obtain a reaction 3 where the same formula applies:

$$\color{\red}{\Delta_rG_3} = \color{\navy}{\Delta_rG^o_3} + \color{\green}{RT\ ln\ Q_3}$$

However, since we can add the free energy equations for reactions 1 and 2, reaction 3 is also represented by:

$$\Delta_rG_1 + \Delta_rG_2 = \Delta_rG^o_1 + \Delta_rG^o_2 + RT\ ln\ Q_1 + RT\ ln\ Q_2$$
$$\color{\red}{\Delta_rG_1 + \Delta_rG_2} = \color{\navy}{\Delta_rG^o_1 + \Delta_rG^o_2} + \color{\green}{RT\ ln\ Q_1 Q_2}$$

Clearly, we have $Q_3 = Q_1 \times Q_2$. Considering the equilibrium conditions of all reactions ($\Delta_rG = 0$), then the reaction quotients are defined as the equilibrium constants $K$, and similarly one finds $K_3 = K_1 \times K_2$.