reaction coordinate, kinetics, equilibrium in example
You have chosen an unusual scheme in that A appears to be an intermediate as well as a reactant. If you use the simpler scheme things are easier and so rather than try to follow yours I have examined the scheme below which is very nonetheless very similar

$$RX\underset{k_{-1}} {\stackrel{k_1}{\leftrightharpoons}}R+X$$
$$R+Y\underset{k_{-2}} {\stackrel{k_2}{\leftrightharpoons}}RY$$

where species RX looses X and is replaced by Y. The intermediate species is R. The overall reaction is $RX+Y=RY+X$ and the equilibrium constant for the two steps are

$$K_1=\frac{k_1}{k_{-1}}=\frac{\mathrm{[R]_e[X]_e}}{\mathrm{[RX]_e}}$$
$$K_2=\frac{k_2}{k_{-2}}=\frac{\mathrm{[RY]_e}}{\mathrm{[R]_e[Y]_e}}$$

and the overall equilibrium constant

$$K=K_1K_2=\frac{k_{1}k_{2}}{k_{-1}k_{-2}}=\frac{\mathrm{[RY]_e[X]_e} }{\mathrm{[RX]_e[Y]_e }}$$

and all the concentrations in square brackets with subscript $e$ are equilibrium values. [This type of equation is true if there are many equilibria one after the other, $\displaystyle K=K_1K_2K_3K_4\cdots=\frac{k_1k_2k_3k_4}{k_{-1}k_{-2}k_{-3}k_{-4}}\cdots$].

In a rate equation approach we can apply a steady state approach to the intermediate species R. The steady state assumes that the rate of change of R is zero;

$$\frac{d[R]}{dt}= k_1[RX]-k_{-1}[R][X]-k_2[R][Y]+k_{-2}[RY]=0$$

from which 

$$ [R]_{ss}= \frac{k_{-2}[RY]+k_1[RX]}{k_{-1}[X]+k_2[Y]}$$

The rate $r$ can be given by

$$\begin{align}
r=-\frac{d[Y]}{dt}=k_2[R][Y]-k_{-2}[RY] &amp;= \frac{k_2(k_{-2}[RY]+k_1[RX])[Y]-k_{-2}[RY](k_{-1}[X]+k_2[Y])}{k_{-1}[X]+k_2[Y]}\\&amp;=\frac{k_1k_2[RX][Y]-k_{-1}k_{-2}[RY][X]}{k_{-1}[X]+k_2[Y]}\\
\end{align}$$

and the numerator is zero when equilibrium concentrations are used making the rate equal to zero also. So this connects the equilibrium to the rate equations.

Initially just after the reactants are mixed the amount of product is very small and $k_1k_2[RX][Y]&gt;&gt;k_{-1}k_{-2}[RY][X] $ and the the rate is 

$$r=\frac{k_1k_2[RX][Y]}{k_{-1}[X]+k_2[Y]}$$

which can be confirmed by experiment. Hope this helps.