Instrumentation Amplifier – Very Low Frequency 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.

You've made an error in the amplification.

\$G = \dfrac{2 \times R_F + R_G}{R_G}\$

With the values you're using this gives

\$G = \dfrac{2 \times 10k\Omega + 25.9k\Omega}{25.9k\Omega} = 1.77\$

If you want a \$G\$ of 750 you'll have to make \$R_G\$ = 26.7\$\Omega\$, so that

\$G = \dfrac{2 \times 10k\Omega + 26.7\Omega}{26.7\Omega} = 750\$