It depends on your (assumed) underlying data generation process.
In general, Weighted Least Squares (WLS) can be used when your data is heteroscedastic but still uncorrelated.
Assume a linear model
$$Y_i = \beta_0 + \beta_1 X_i + \epsilon_i \tag{1}$$
If you assume $var(\epsilon_i) = \sigma^2$, i.e. the error terms are homoscedastic, OLS is the best linear unbiased estimator (BLUE). However, if you allow errors to be heteroscedastic, we have $var(\epsilon_i) = \sigma_i^2$, so the variance of residuals depends on the specific observation. However, you can rewrite the latter model as:
$$var(\epsilon_i) = \sigma_i^2 = \sigma^2 \cdot d_i \tag{2}$$
,so you can account for heteroscedasticity by assuming an overall constant error variance (just like OLS), but weighting each error term with a factor $d_i$. If you would divide $\epsilon_i$ by $d_i$, as in $\theta_i = \frac{\epsilon_i}{\sqrt{d_i}}$, you obtain
$$var(\theta_i) = var \left( \frac{\epsilon_i}{\sqrt{d_i}} \right)= \frac{\sigma_i^2}{d_i} = \sigma^2 = const \tag{3}$$
,which makes OLS applicable again. In fact, assuming (2), WLS is just OLS with a transformed model by dividing any observation by $\sqrt(d_i)$.
So how is the underlying weighting $w_i$ for any observation $x_i$ in the least square algorithm? In the case of OLS, we have $w_i \propto X_i$, wheres in WLS, each observation weight is proportional to $X_i / \sqrt{d_i}$.
In summary, for $d_i$ as the market capitalization of a firm, if you assume for the residual variance that $var(\sigma_i^2) = \sigma^2 \cdot d_i$ holds, i.e. the error variance is proportional to the market capitalization, you have to weight each observation $X_i$ with $\sqrt{d_i}$.