Error in this explanation of instrumentation amplifier?
You're right, this is wrong. If the output of the buffer amplifier would be equal to \$V_1\$ then there wouldn't be any voltage difference over R2, because the opamp will keep the inverting input at \$V_1\$ as well. So there wouldn't be any current through either R2. Yet there's a voltage difference \$V_2\$ - \$V_1\$ between both inverting inputs, causing a current through R1. Since no current flows in the opamp's inputs this would be a violation of KCL (Kirchhoff's Current Law).
So, the buffer amplifiers are not voltage followers and \$V_a\$ is not equal to \$V_1\$. The equation further on the page also mentions the factor \$\dfrac{R4}{R3}\$ a bit too early. It should read    


  \$\dfrac{V_b - V_a}{V_2 - V_1} = 1 + \dfrac{2 R2}{R1}\$


This follows from the current flowing between \$V_a\$ and \$V_b\$:  


  \$I = \dfrac{V_b - V_a}{R2 + R1 + R2}\$  


and since the same current flows through R1:  


  \$I = \dfrac{V_2 - V_1}{R1}\$  


hence  


  \$\dfrac{V_b - V_a}{R1 + 2 R2} = \dfrac{V_2 - V_1}{R1}\$


The factor \$\dfrac{R4}{R3}\$ is the amplification by the differential amplifier stage.