Sum of Ito processes A stochastic processes $X_t$ is called an Ito process if $X_t$ is of the form
$$X_t = X_0 + \int_0^tu_sdB_s + \int_0^t v_sds,$$
where $B_s$ is a Brownian motion, and $u_s, v_s$ are square integrable and adapted to the filtration generated by $B_s$.

Now, suppose I have 2 Ito processes driven by two independent Brownian motions, say
$$X_t = X_0 + \int_0^tu_sdB_s + \int_0^t v_sds,$$
$$\tilde{X}_t = \tilde{X}_0 + \int_0^t\tilde{u}_sd\tilde{B}_s + \int_0^t \tilde{v}_sds,$$
where $B_s$ and $\tilde{B}_s$ are two independent Brownian motions.

We may regard $(X_t,\tilde{X}_t)$ as a two dimensional Ito processes, and by Ito's lemma $X_t + \tilde{X_t}$ is also an Ito process (one dimensional).

My question is, how can we write $X_t + \tilde{X}_t$ as an Ito process?
Specifically, I want to produce the following form
$$X_t + \tilde{X}_t = X_t + \tilde{X}_0 + \int_0^t\bar{u}_sd\bar{B}_s + \int_0^t \bar{v}_sds.$$
So that, $\bar{B}_s$ may depend on $B_s$ and $\tilde{B}_s$.
It is quite clear that $\bar{v}_t = v_t + \tilde{v}_t$. My question really concern $\bar{B}_t$ and $\bar{v}_t$.
I am sorry I don't have the answer but I just got interested in this question too. I think "m7e" is not responding something satisfactory and I came across equations where the author seems to sum Itô processes with different driving Brownian motions. Given his calculus, it looks like the following holds:

if $$dX_t = a_tdW_t^{(a)} +  b_tdW_t^{(b)} $$
then by definition $X_t-X_0=\int_0^ta_sdW_s^{a} + \int_0^tb_sdW_s^{b} $, and the author writes $$dX_t = a_tdW_t^{(a)} +  b_tdW_t^{(b)} = \sqrt{a_t^2+b_t^2} dW_t $$
where $W_t$ is brand new Wiener process.
This means that:
$$\int_0^ta_sdW_s^{a} + \int_0^tb_sdW_s^{b} = \int _0^t \sqrt{a_s^2+b_s^2}dW_s $$ 
and $W_t$ is only one-dimensional. I'll come back to you if I find the right theorem for this!
Usually one allows the driving Brownian motion to be multidimensional, and, correspondingly, $u$ to be vector-valued. With this convention you simply set $\bar B=(B,\tilde B)$ and $\bar u=(u,\tilde u)$.