I'm aware of several related stack questions, but my case is a bit different because I assume that the system of SDEs is multiplied by 1-dimensional increment $dW$.
Suppose $\sigma$ is an $n \times n$ matrix and $dW$ is a 1-dimensional Wiener increment. The system of Stratonovitch SDEs is: $$dX = \sigma X \circ dW.$$ I'm unsure how to derive the correction term to convert the system to the Ito form. So far, I think the Ito system of SDEs should be $$dX = \sigma X dW + \frac{1}{2}\sigma^2 Xdt.$$
This is true.
In the one-dimensional case the transformation between the two types of integrals is, as we know, $$ \underbrace{\int_0^tY_s\circ\,dW_s}_{\text{Stratonovich}}=\underbrace{\int_0^tY_s\,dW_s}_{\text{Ito}}+\tfrac{1}{2}\langle W,Y\rangle_t\,. $$ In particular,
In your case $$ Y_i(t)=(\sigma X(t))_i=\sum_{j=1}^n\sigma_{ij}X_j(t)\,,\quad\langle W,Y_i\rangle_t=\sum_{j=1}^n\sigma_{ij}\langle W,X_j\rangle_t\,. $$ Using the above bullet point and bilinearity of the covariation, \begin{align} \langle W,X_j\rangle_t&=\left\langle W,\int_0^.\sum_{k=1}^n\sigma_{jk}X_k(s)\circ\,dW_s\right\rangle=\sum_{k=1}^n\sigma_{jk}\left\langle W,\int_0^.X_k(s)\,dW_s\right\rangle_t\\[2mm] &=\sum_{k=1}^n\sigma_{jk}\int_0^tX_k(s)\,ds\,. \end{align} This shows \begin{align} dX_i(t)=\sum_{j=1}^n\sigma_{ij} X_j(t)\circ dW_t=\sum_{j=1}^n\sigma_{ij} X_j(t)\,dW_t+\frac12\sum_{j,k=1}^n\sigma_{ij}\,\sigma_{jk}X_k(t)\,dt\,. \end{align} In shorter matrix/vector notation this is the formula you wrote at the end of OP.
