TLDR The difference between the two situations is whether you use $\sqrt{\frac{s_a^2}{n_a} + \frac{s_b^2}{n_b}}$ or $\sqrt{\frac{s^2}{n_a} + \frac{s^2}{n_b}}$.
In the second case the estimate of the variance of the populations $a$ and $b$ is coupled based on the assumption that the populations have equal variance.
The nasty looking formula in the second case stems from deriving the pooled sample deviation $s$ based on the individual sample deviations $s_a$ and $s_b$.
The formula's might become more intuitive when you consider the sum of the squared residuals from which the sample standard deviation is derived.
$$ \sum_{i=1}^n {r_i^2} = \sum_{i=1}^n (x_i - \bar{x})^2$$
This $r_i^2$ is a sum of $n$ terms but effectively equivalent to the sum of $n-1$ squared independent normal distributed variables with variance $\sigma$. (See for instance Why are the residuals in $\mathbb{R}^{n-p}$?)
The distribution of squared independent normal distributed variables with standard deviation $\sigma$ follows a gamma distribution* with shape parameter $n-1$ and scale parameter $\sigma^2$ and will have a mean of $(n-1)\sigma^2$. So if we divide by $n-1$ then we have an unbiased estimate of the variance.
$$s^2 = \hat{\sigma^2} = \frac{1}{n-1} \sum_{i=1}^n {r_i^2}$$
And $s$ is the corrected sample estimate of the standard deviation
$$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n {r_i^2}}$$
Now, if we have more residual terms because we sampled two population where we assume that the two populations have the same variance (that is what the pooling does). Then we get simply a sum of those residual terms and now it is a gamma distributed variable that is equivalent to a sum of $(n_a - 1)+(n_b -1)$ squared normal distributed variables.
$$s = \sqrt{\frac{1}{(n_a-1)+(n_b-1)} \left(\sum_{i=1}^{n_a} {r_{a,i}^2} + \sum_{i=1}^{n_b} {r_{b,i}^2}\right)}$$
where $n_a$ and $n_b$ are the sizes of the two samples and $r_{a,i}$ and $r_{b,i}$ the residual terms in the two samples.
If instead of the sum of squared residuals you use the corrected sample standard deviations
$$\sum_{i=1}^{n_a} {r_{a,i}^2} = s_a^2 (n_a -1)\\
\sum_{i=1}^{n_b} {r_{a,i}^2} = s_b^2 (n_b -1)$$
then you get
$$s = \sqrt{\frac{s_a^2 (n_a -1) +s_b^2 (n_b -1)}{(n_a-1)+(n_b-1)}}$$
The additional term $\sqrt{\frac{1}{n_a}+\frac{1}{n_b}}$ is to convert from an estimate about the variance/deviation of the population to the variance/deviation of the sample mean or the difference between two sample means. The estimate of the variance of the one mean will be $s^2/n_a$ and the other $s^2/n_b$. The estimate for the variance of the difference is the sum of those two.
*Some readers might be more familiar with the $\chi^2$ distribution which is a special case of the gamma distribution, but with a scale equal to 1.
