Taylor expansion to show that for Stratonovich stochastic calculus the chain rule takes the form of the classical one How can I show with a heuristic argument based on a Taylor expansion that for Stratonovich stochastic calculus the chain rule takes the form of the classical (Newtonian) one? 
Concerning Ito calculus the fact that $dX^2 = dt$ results via a Taylor expansion in Ito's lemma -- this fact should stay the same with Stratonovich, but it should somehow cancel out in there -- I just don't know how....
 A: In the following, $W_t$ is a Wiener process with increments having mean zero and standard deviation 1.
Basically, the Stratonovich formula does the following in terms of the Itô formula:
$$f(W_t) \circ dW_t = \frac{f(W_t+dW_t)+f(W_t)}{2}dW_t \; ,$$
which can be rewritten as
$$f(W_t) \circ dW_t = \frac{f(W_t+dW_t)-f(W_t)}{2}dW_t + f(W_t)dW_t \; .$$
Using Taylor expansion on the first term and Itô calculus rules, this can be simplified to
$$f(W_t) \circ dW_t = \frac{f'(W_t)}{2}dt + f(W_t)dW_t \; .$$
Now, we can replace $f$ with $f'$ in the formula to give
$$f'(W_t) \circ dW_t = \frac{f''(W_t)}{2}dt + f'(W_t)dW_t \; .$$
But you can easily check that the right hand side is nothing but $df(W_t)$ according to Itô calculus rules, therefore
$$df(W_t) = f'(W_t) \circ dW_t \; .$$
Which is just a special case of the chain rule. I guess from here on out you can generalize the argument.
A: I have published a pedagogical paper which, among other things, contains the Taylor expansion for the Stratonovich case (p. 14):
$$f(x_{t+1})\approx f(x_t)+f' \left(x_t+\frac{x_{t+1}-x_t}{2}\right)(x_{t+1}-x_t)$$
$$=f(x_t)+f' \left(\frac{x_{t+1}+x_t}{2}\right)(x_{t+1}-x_t)$$
For the integral notation, we have to make clear that we are now talking about a different integral.
To that end we chose a slightly different notation with a circle in front of the integrator. This circle
makes it clear that we use the mid-point for the Stratonovich integral: 
$$f(X_T)-f(X_0)=\int_0^Tf'(X_t) \circ dX_t$$
As one can see the second term disappears in this case. The reason for this is that when one takes the right-hand point one gets a result like the Ito integral but with a negative correction term. The Stratonovich integral is just the arithmetic mean of the two and therefore loses the correction term.
And when one compares this in differential notation to the classic chain rule one can see, that we have really achieved what we wanted: they both have the same form:
$$df(X_t)=f' \left(X_{t+\frac{1}{2}} \right)dX_t$$
(Strictly speaking this formula would not make sense since it would not be clear what was meant by $t+\frac{1}{2}$ as a subscript, so it is given here for illustrative purposes only.)
More details can be found in the paper:
von Jouanne-Diedrich, Holger, Ito, Stratonovich and Friends (May 18, 2017). Available at SSRN: https://ssrn.com/abstract=2956257 or http://dx.doi.org/10.2139/ssrn.2956257
