Can we apply Weighted Least Squares to remove heteroscedasticity when $\sigma_i^2=\sigma^2x_i^2$ for a regression like $Y_i=\beta_0+\beta_1x_i+u_i$? To be honest, the reason why I'm asking this question is that the expression I end up with looks a little bit weird to me, which is the following,
$Y_i/x_i=\beta_0/x_i+\beta_1x_i/x_i+u_i/x_i$
$Y_i/x_i=\beta_0/x_i+\beta_1+u_i/x_i$
 A: That's correct -- you can apply weighted least squares. However, what you've done by dividing through by $x$ is allow yourself to use an unweighted fit to a transformed model in order to obtain correct WLS estimates.
Here's an example showing that this works "as advertised" when the model is right -- as long as your x's are strictly positive.
We start on the left with data from a model with population intercept 3 and slope 1.4, with noise variance proportional to $x^2$. Then dividing through by x we convert to plotting y/x vs 1/x (and the intercept and slope swap roles). This stabilizes the variance, making unweighted least squares suitable:
 
We obtain the fitted line (in dark orange on the right hand figure) in R using your transformed equation by doing:
 lm(I(y/x)~I(1/x))  # no weights here (but intercept and slope swap)

or we could also have done it directly by weighted least squares:
 lm(y~x, weights=1/x^2)  # this is weighted least squares

either way we get the same estimates. 
The fitted line looks good; and looks fine when we transfer it back:

An unweighted line also fits well (it is only a bit less accurate on average) but the estimated standard errors (and t-values and p-values and confidence intervals etc) would be wrong.
