The following is the first step in the proof of Morse's Lemma:
Without loss of generality, we can assume that $\overline x$ is the origin in $\mathbb R^N$ (in the sense that the chart takes $\overline x$ to the origin) and that $f(\overline x) = f(0) = 0$ (trivial). Note that we can write, for $x$ in a neighborhood of the origin, \begin{align*} f(x) & = \int_0^1 \sum_{i = 1}^N \frac{\partial f}{\partial x_i} (tx) x_i \ dt \\ & = \sum_{i,j = 1}^N \left(\int_0^1 \int_0^1\frac{\partial^2f}{\partial x_i \partial x_j} (stx) \ ds \ dt\right)x_ix_j \\ & = \sum_{i, j = 1}^N h_{ij}(x) x_i x_j. \end{align*}
Then, Milnor argues that we can assume $h_{ij} = h_{ji}$ by taking $\overline h_{ij} = \frac 12(h_{ij} + h_{ji})$ if necessary. My question is, isn't $h_{ij} = h_{ji}$ already, since the function $f$ is smooth?
Milnor applies the original lemma to the functions $g_i(x)=\displaystyle\int_0^1 \dfrac{\partial f}{\partial x_i}(tx)\,dt$ rather than trying to give the formula explicitly. Thus, he is not a priori guaranteed any symmetry.
I suggest that you did not correctly differentiate under the integral sign when you gave your explicit formula. Indeed, let's try a simple example, even with $n=1$. Take $f(x)=x^2$. Then $g_1(x)=\int_0^1 f'(tx)\,dt = \int_0^1 2tx\,dt = x$, as of course it should be. Now your formula gives $\int_0^1\int_0^1 f''(stx)\,ds\,dt = \int_0^1\int_0^1 2\,ds\,dt = 2$, which does not recover the function $f$. We're off by a numerical factor. You need to go back and compute the derivative $\dfrac{\partial g_i}{\partial x_j}$ more carefully. (Examples are your friends!!)
That said, I think that when you make your explicit integral formula correct, you will in fact end up proving what you believed in the first place, i.e., that $h_{12}=h_{21}$ when $f$ is $C^2$.