Short answer: There is no exact degrees-of-freedom because the variance estimator in this test does not follow an exact chi-squared distribution.
Longer answer: The Welch T-test gives an approximate solution to the Fisher-Behrens problem (comparing the means of two samples with different variances). It uses the studentised test statistic:
$$T = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{S_1^2/N_1 + S_2^2/N_2}}.$$
The denominator in this test statistic is the square-root of an estimator of the mean difference:
$$\hat{\mathbb{V}}(\bar{X}_1-\bar{X}_2) = \frac{S_1^2}{N_1} + \frac{S_2^2}{N_2} \sim \frac{\chi_{N_1-1}^2}{N_1} \cdot \sigma_1^2 + \frac{\chi_{N_2-1}^2}{N_2} \cdot \sigma_2^2.$$
This quantity is a weighted sum of independent chi-squared random variables. Its exact distribution is quite complicated (and is best represented through its moment generating function), but it is not an exact chi-squared distribution.
The test uses the Welch-Satterwaite approximation, which approximates the distribution of this quantity by a single scaled chi-squared distribution. In this approximation the degrees-of-freedom formula arises as the best approximation of the chi-squared distribution to the true distribution of this quantity. Without this approximation to the chi-squared distribution there is no single exact degrees-of-freedom. Instead, the exact distribution is of a weighted sum of chi-squared random variables with the above weights and degrees-of-freedom.