Why May K Expressions Be Manipulated As Such?
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 $K_3 = K_1 \times K_2$.