The following answer is not complete and only give some intuitive grasp on Fisher's fundamental theorem of Natural Selection. A better devlopment can be found in Ewen's book
Let's first define what is the Additive Genetic Variance
Consider a quantitative character that is determined entirely by a locus $A$ which two alleles are $A_1$ and $A_2$. the measurement $m$ of this quantitative character for an individual is given by their genotypes so that the genotypes $A_1A_1$, $A_1A_2$ and $A_2A_2$ have measurements $m_{11}$, $m_{12}$ and $m_{22}$ respectively. Suppose that random mating obtains with respect to this character and that the frequencies of $A_1A_1$, $A_1A_2$ and $A_2A_2$ are $x^2$, $2x(1-x)$ and $(1-x)^2$, respectively. Then, the mean value $\bar m$ of this measurement is
$$\bar m = x^2m_{11} + 2x(1-x)m_{12} + (1-x)^2m_{22}$$
and the variance in the measurement is
$$\sigma ^2 = x^2(m_{11} - \bar m) + 2x(1-x) (m_{12} - \bar m) + (1-x)^2 (m_{22} - \bar m)$$
The covariance between the fathers and the sons (assuming no change in allele frequency) is
$$x(1-x)\left(xm_{11} + (1-2x)m_{12} - (1-x)m_{22}\right)^2$$
The correlation between the fathers and the sons is found by dividing the covariance by the variance (since the variance of fathers equal the variance of sons) is
$$\frac{x(1-x)\left(xm_{11} + (1-2x)m_{12} - (1-x)m_{22}\right)^2}{\sigma ^2}$$
which can be decomposed into dominance and additive variance
$$\sigma _A ^2 = 2x(1-x) (xm_{11} + (1-2x)m_{12} - (1-x)m_{22})^2$$
$$\sigma _D ^2 = x^2(1-x)^2 (2xm_{12} - m_{11} - m_{22})^2$$
Replacing the measurement by the fitness, you get the additive genetic variance in fitness.
$$\sigma _A ^2 = 2x(1-x) (xw_{11} + (1-2x)w_{12} - (1-x)w_{22})^2$$
$$\sigma _D ^2 = x^2(1-x)^2 (2xw_{12} - w_{11} - w_{22})^2$$
Now, the mean fitness in the population is given by (as already said)
$$\bar w = x^2w_{11} + 2x(1-x)w_{12} + (1-x)^2w_{22}$$
Using Wright-Fisher equation, the change in mean fitness is
$$\Delta \bar w = 2x(1-x) (xw_{11} + (1-2x)w_{12} - (1-x)w_{22})^2 \cdot (w_{11}x^2 + (w_{12} + \frac{w_{11} + w_{22}}{2})x(1-x) + w_{22}(1-x)^2)\bar w ^{-2}$$
which can be approximated
$$\Delta \bar w ≈ 2x(1-x) (xw_{11} + (1-2x)w_{12} - (1-x)w_{22})^2 = \sigma _A ^2$$
Reference