Why doesn't the TOST equivalence testing procedure use non-central $t$ distribution to determine the $p$ value?

I'm looking at Lakens' (2017) primer and he tests the hypothesis that there is a difference between two groups $X_1$and $X_2$of magnitude $\Delta=E[X_1]-E[X_2] \neq0$ by subtracting $\Delta$ from the difference between the sample means $d=M_1-M_2$ and then he computes the $t$ value and $p$ value based on this difference score with a procedure analogous to Welch t-test (Eqs 3 and 4 on p. 357).

This looks wrong to me. IMO to determine the p value, one should be using non-central $t$ distribution with non-centrality parameter based on $\Delta$ and t value based on $d$. This is because under the hypothesis, $d$ and hence the test statistic $t=d/h$ (where we would substitute $h=\sqrt{s_1^2/n_1+s_2^2/n_2}$ following Welch) does not have a symmetric distribution because $E[X_1]-E[X_2]=\Delta\neq 0$. Subtracting $\Delta$ from $d$ does merely shift the distribution, but it can't make it symmetric, since $\Delta$ is not a random variable.

Another CrossValidated answer claims that

Using the zero centered T distribution here would test the hypothesis assuming that X1 - X2 -3 is symmetric about zero.`

(X1-X2-3 corresponds to $X_1-X_2-\Delta$). I don't see how it is possible to obtain such symmetric test statistic distribution unless $E[X_1]=E[X_2]$ which under the hypothesis is not the case ($E[X_1]-E[X_2]=\Delta\neq 0$).

So, which is the correct procedure to conduct equivalence test?

Literature

Lakens, D. (2017). Equivalence tests: a practical primer for t tests, correlations, and meta-analyses. Social Psychological and Personality Science, 8(4), 355-362.

What you are proposing does not appear to be any different from what actually occurs in this test. Remember that in a classical hypothesis test, the p-value is calculated using the null distribution of the test statistic. This calculation assumes that the null hypothesis is true. You suggest that the calculation of the p-value should use the non-central T-distribution with non-centrality parameter $\Delta$. However, the null hypothesis for this test is that $\Delta = 0$, so under that condition the non-central T-distribution simplifies down to the standard (centralised) T-distribution.