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In the case of two independent samples, the formula for standard error of the difference in means is given by :

$$\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$$

Even though we are talking about a difference, why are the standard errors of each sample means are added together?

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The standard error is an estimation of the standard deviation of the statistic of your interest.

In your case, the statistic at hand is $\Delta = \bar{X}_1 - \bar{X}_2$, where $\bar{X}_i$ is the sample mean of an i.i.d. sample $\mathcal{D}_i := \{X_{i, 1}, \ldots, X_{i, n_i}\} \sim(\mu_i, \sigma_i^2)$, $i = 1, 2$. Under the additional assumption that $\mathcal{D}_1$ and $\mathcal{D}_2$ are independent, the standard deviation of $\Delta$ is clearly \begin{align*} \sigma := \sqrt{\operatorname{Var}(\Delta)} = \sqrt{\operatorname{Var}(\bar{X}_1) + \operatorname{Var}(\bar{X}_2)} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}. \tag{1}\label{1} \end{align*} In the above calculation, you are seeing a "$+$" instead of "$-$" in the right hand side expression because $\operatorname{Var}(X - Y) = \operatorname{Var}(X) + \operatorname{Var}(Y)$ rather than "$\operatorname{Var}(X - Y) = \operatorname{Var}(X) - \operatorname{Var}(Y)$" (again, given that $X$ and $Y$ are independent). This is a probabilistic rule, and actually has nothing to do with statistical inference.

Now if you estimate $\sigma_i^2$ in $\eqref{1}$ by corresponding sample variance $s_i^2$, you then retrieved the standard error expression as you listed.

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