Why does the McNemar's test use $chi^{2}$ and not the normal distribution?

A close-to-intuitive answer:

Take a closer look at the formula for the McNemar test, given the table

      pos | neg
----|-----|-----
pos |  a  |  b
----|-----|-----
neg |  c  |  d

The McNemar statistic M is calculated as:

$$ M = {(b-c)^2 over b+c} $$

The definition of a $chi^2$ distribution with k degrees of freedom is that it consists of the sum of squares of k independent standard normal variables. if the 4 numbers are large enough, b and c, and thus b-c and b+c can be approximated by a normal distribution. Given the formula for M, it's easily seen that with large enough values M will indeed follow approximately a $chi^2$ distribution with 1 degree of freedom.

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EDIT : As onstop rightfully indicated, the normal approximation is in fact completely equivalent. That's rather trivial given the argument using the approximation of b-c by the normal distribution.

The exact binomial version is also equivalent to the sign test, in the sense that in this version the binomial distribution is used to compare b to $Binom(b+c,0.5)$. Or we can say that under the null hypothesis the distribution of b can be approximated by $N(0.5times(b+c),0.5^2times(b+c)$.

Or, equivalently:

$$frac{b-(frac{b+c}{2})}{frac{sqrt{b+c}}{2}}sim N(0,1)$$

which simplifies to

$$ frac{b-c}{sqrt{b+c}}sim N(0,1)$$

or, when taken the square on both sides, to $M sim chi^2_1$.

Hence, the normal approximation is used. It is the same as the $chi^2$ approximation.