Suppose you start $\{x_i\}$ with mean $m_1$ and non-zero standard deviation $s_1$ and you want to arrive at a similar set with mean $m_2$ and standard deviation $s_2$.
Then multiplying all your values by $\frac{s_2}{s_1}$ will give a set with mean $m_1 \times \frac{s_2}{s_1}$ and standard deviation $s_2$.
Now adding $m_2 - m_1 \times \frac{s_2}{s_1}$ will give a set with mean $m_2$ and standard deviation $s_2$.
So a new set $\{y_i\}$ with $$y_i= m_2+ (x_i- m_1) \times \frac{s_2}{s_1} $$ has mean $m_2$ and standard deviation $s_2$.
You would get the same result with the three steps: translate the mean to $0$, scale to the desired standard deviation; translate to the desired mean.
Think of the difference like any other statistic that you are collecting. These differences are just some values that you have recorded. You calculate their mean and standard deviation to understand how they are spread (for example, in relation to 0) in a unit-independent fashion.
The usefulness of the SD is in its popularity -- if you tell me your mean and SD, I have a better understanding of the data than if you tell me the results of a TOST that I would have to look up first.
Also, I'm not sure how the difference and its SD relate to a correlation coefficient (I assume that you refer to the correlation between two variables for which you also calculate the pairwise differences). These are two very different things. You can have no correlation but a significant MD, or vice versa, or both, or none.
By the way, do you mean the standard deviation of the mean difference or standard deviation of the difference?
Update
OK, so what is the difference between SD of the difference and SD of the mean?
The former tells you something about how the measurements are spread; it is an estimator of the SD in the population. That is, when you do a single measurement in A and in B, how much will the difference A-B vary around its mean?
The latter tells us something about how well you were able to estimate the mean difference between the machines. This is why "standard difference of the mean" is sometimes referred to as "standard error of the mean". It depends on how many measurements you have performed: Since you divide by $\sqrt{n}$, the more measurements you have, the smaller the value of the SD of the mean difference will be.
SD of the difference will answer the question "how much does the discrepancy between A and B vary (in reality) between measurements"?
SD of the mean difference will answer the question "how confident are you about the mean difference you have measured"? (Then again, I think confidence intervals would be more appropriate.)
So depending on the context of your work, the latter might be more relevant for the reader. "Oh" - so the reviewer thinks - "they found that the difference between A and B is x. Are they sure about that? What is the SD of the mean difference?"
There is also a second reason to include this value. You see, if reporting a certain statistic in a certain field is common, it is a dumb thing to not report it, because not reporting it raises questions in the reviewer's mind whether you are not hiding something. But you are free to comment on the usefulness of this value.
Best Answer
I am assuming you do not have the observed data, and just have the means and standard deviations. Let the data from the left axilla be $X_1, X_2, \dots, X_n$, and let the data from the right axilla be $Y_1, Y_2, \dots, Y_n$. I am assuming that you have equal sample sizes.
(Before I go on, I must warn you that your pooled/combined sample is no longer independent, because you have two data points on each individual. This is important if you are going to use the standard deviations to test a hypothesis.)
You are given the quantities $\bar{X}_n = 36.4$ and $sd(X_1) = 0.2$ and $\bar{Y}_n = 36.7$ and $sd(Y_1) = .18$.
(Here $\bar{X}_n$ denotes sample mean of $X$s similarly for $Y$).
If you were to pool all the data, you would have $X_1, \dots, X_n, Y_1, \dots, Y_n$. Let these be noted by $Z_1, \dots, Z_{2n}$. Then mean of $Z$, $$\bar{Z}_{2n} =\dfrac{X_1 + X_2 + \dots + X_n + Y_1 +Y_2 + \dots Y_n}{2n} = \dfrac{\bar{X}_n + \bar{Y}_n}{2} = 36.55$$
The standard deviation can be calculated using pooled estimators.
$$sd(Z_1)^2 = \dfrac{(n-1)sd(X_1)^2 + (n-1)sd(Y_1)^2}{2(n-1)} = \dfrac{sd(X_1)^2 + sd(Y_1)^2}{2}.$$
Thus the standard deviation is
$$sd(Z_1) = \sqrt{\dfrac{sd(X_1)^2 + sd(Y_1)^2}{2}}. $$
Like I mentioned before, since your sample is not independent, note that
$$sd(\bar{Z}_{2n})^2 \ne \dfrac{sd(Z_1)^2}{2n}$$.