From Wikipedia the multidimensional Ito lemma is:
If $\mathbf{X}_t = (X^1_t, X^2_t, \ldots, X^n_t)^T$ is a vector of Itō processes such that $d\mathbf{X}_t = \boldsymbol{\mu}_t\, dt + \mathbf{G}_t\, d\mathbf{B}_t$ for a vector $\boldsymbol{\mu}_t$ and matrix $\mathbf{G}_t$, Itō's lemma then states that
$$df(t,\mathbf{X}_t) = \left\{ \frac{\partial f}{\partial t} + \left (\nabla_\mathbf{X} f \right)^T \boldsymbol{\mu}_t + \frac{1}{2}\text{Tr}\left[ \mathbf{G}_t^T \left( H_\mathbf{X} f \right) \mathbf{G}_t \right] \right\} dt + \left (\nabla_\mathbf{X} f \right)^T \mathbf{G}_t\, d\mathbf{B}_t$$
Now let $\mathbf{X}_t = ((X_1)_t, (X_2)_t)$, where $X_i = \mu_i \; dt + \sigma_i \; d(B_i)_t$. Then matrix $\mathbf{G}$ is just $\begin{pmatrix} \sigma_1 & 0 \\ 0 & \sigma_2 \end{pmatrix}$, $\nabla_\mathbf{X} f$ is $(x_2, x_1)$, and $H_\mathbf{X} f = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. Define $f(t, x_1, x_2) = x_1 x_2$. Then from Ito, we should have
\begin{eqnarray*} d(X_1 X_2) &=& (X_2, X_1) \cdot (\mu_1, \mu_2) + (X_2 \sigma_1, X_1 \sigma_2) \cdot (dB_1, dB_2) \\ &=& X_2 (dX_1) + X_1 (dX_2) \end{eqnarray*}
However, this is not the correct answer. We should have $$d(X_1 X_2) = X_2 (dX_1) + X_1 (dX_2) + (dX_1)(dX_2)$$ Why am I missing the $(dX_1)(dX_2)$ term above?
The cross quadratic variation process $[X_1,X_2]$ is $0$. That is why it is not showing up.
You may wonder why $[X_1,X_2]$ is $0$.
$$[X_1,X_2]_t = [\int_0^t\mu_1\,ds + \int_0^t\sigma_1\,dB_{1,s},\int_0^t\mu_2\,ds + \int_0^t\sigma_2\,dB_{2,s}]_t $$
You need a bunch of facts to show that this is $0$.
$1$-) Bilinearity of quadratic variation
$2$-) Cross quadratic variation of a continuous process with bounded variation and another cadlag process is $0$.
$3$-) Cross quadratic variation of two independent Brownian motions is $0$.