Do Cohen's d and Hedges' g apply to the Welch t-test? How do you go about calculating effect size for the Welch t-test?
 A: Depends on what is meant by "effect size" here. Welch's t-test is used to test the null hypothesis $\mu_1 = \mu_2$ when we cannot or don't want to assume that the variances are homoscedastic within the two groups.
So what is a good effect size measure to go along with this test? Obviously, it should express how different the means are in the sample. So, we could just compute the mean difference (which is an effect size measure) and there are no difficulties in computing it in this case. It's just $$y = \bar{x}_1 - \bar{x}_2,$$ whose variance can be estimated with $$\mbox{Var}[y] = \frac{SD^2_1}{n_1} + \frac{SD^2_2}{n_2}.$$
However, with "effect size", people often mean some kind of standardized measure (like Cohen's d) and the title makes it clear that this is what the asker is after. There are at least two possibilities here.
The first is to standardize the mean difference by the standard deviation from one of the two groups (e.g., if one group is a control and the other a treatment group, then we could standardize using the control group SD). So, we compute $$y = \frac{\bar{x}_1 - \bar{x}_2}{SD_1},$$ whose variance can be estimated with $$\mbox{Var}[y] = \frac{1}{n_1} + \frac{SD^2_2/SD^2_1}{n_2} + \frac{y^2}{2n_1}.$$
Alternatively, if it does not make sense to choose one of the two SDs for the standardization, then we could proceed as suggested by Bonett (2008) and standardize based on the average SD. This is computed by averaging the two variances and then taking the square-root, that is, let $$\overline{SD} = \sqrt{\frac{SD^2_1 + SD^2_2}{2}}.$$ Then we compute $$y = \frac{\bar{x}_1 - \bar{x}_2}{\overline{SD}},$$ whose variance can be estimated with $$\mbox{Var}[y] = \frac{y^2}{8(\overline{SD})^4} \left(\frac{SD_1^4}{n_1-1} + \frac{SD_2^4}{n_2-1}\right) + \frac{SD_1^2 / \overline{SD}^2}{n_1-1} + \frac{SD_2^2 / \overline{SD}^2}{n_2-1}.$$
A: You don't. You calculate the effect size using the data, irrespective of the kind of T-test you used. One package in R is  effsize.  
d <- cohen.d (y ~ factor (x), hedges.correction = TRUE)

You can also subtract mu2 from mu1 and divide that difference by the average of the standard deviations. This will result in Cohen's d.
