Derivation of an equality in Michaelis–Menten kinetics Enzymatic action may be described as follows: 

$$\ce{Enzyme + Substrate &lt;=&gt;[k_1] ES complex -&gt;[k_\mathrm{2}] Enzyme + Product}$$

The initial rate of enzyme-catalyzed reactions can be described by the Michaelis-Menten equation: 

$$\mathrm{rate} = \frac{V_\mathrm{max}[\ce{S}]}{K_\mathrm{M} + [\ce{S}]} = \frac{k_\mathrm{cat}[\ce{E}][\ce{S}]}{K_\mathrm{M} + [\ce{S}]}$$

where $V_\mathrm{max}$ is the maximum rate, $[\ce{S}]$ the substrate concentration, $[\ce{E}]$ is the enzyme concentration, $K_\mathrm{M}$ is the Michaelis constant and $k_\mathrm{cat}$ is the number of catalytic cycles per second. 

It is known that 

$$V_\mathrm{max} = k_2[E]_0$$

and by inspecting the equation above we can deduce that 

\begin{align}
V_\mathrm{max} &amp;= k_\mathrm{cat}[\ce{E}]\\
\implies k_2[\ce{E}]_0 &amp;= k_\mathrm{cat}[\ce{E}]
\end{align}

How, and why, does this equality hold? 

How do we know when to use ${[\ce{E}]_0}$ and $[\ce{E}]$ in rate equations, and what are the implications of using either? 

${[\ce{E}]_0}$ refers to the enzyme concentration at the start of the reaction, and [E] refers to the concentration of enzyme at any point in time during the course of the reaction.
Your equation


  $$\ce{Enzyme + Substrate &lt;=&gt;[k_1] ES complex -&gt;[k_2] Enzyme + Product}$$


contains only two rate constants, $k_1$ and $k_2$, but not $k_{cat}$ you refer to.

The correct schema for the Michaelis-Menten kinetics would be

$$\ce{Enzyme + Substrate &lt;=&gt;[k_1][k_{-1}] ES complex -&gt;[k_2 = k_{cat}] Enzyme + Product}$$

Note that $k_{cat} = k_2$ is the rate constant for the decay of the ES complex in the second step.

From this you can derive the Michaelis-Menten rate law as described in the Wikipedia article you refer to.

$[E]_0$ is the initial concentration of the enzyme. During the reaction the enzyme exists as free enzyme ($[E]$) and bound in the enzyme substrate complex $[ES]$. These concentrations are related by the equation $[E]_0 = [E] + [ES]$.