Can two elements always belong to the same Laurent series field?

Yes.

The missing ingredient here is the field $\mathbb F_q(x,y)$. The field $\mathbb F_q((\frac{1}{T})) (x,y)$ contains $\mathbb F_q(x,y)$, is complete (being a finite extension of $\mathbb F_q((\frac{1}{T}))$), and $\mathbb F_q(x,y)$ is dense in it (since $\mathbb F_q(T)$ is dense in $\frac{1}{T}$ and $\mathbb F_q(x,y)$ generates $\mathbb F_q((\frac{1}{T})) (x,y)$ as a $\mathbb F_q((\frac{1}{T})) $-vector space), so $\mathbb F_q((\frac{1}{T})) (x,y)$ is the completion of $\mathbb F_q(x,y)$ with respect to a valuation $v$ given by restricting the valuation on $\mathbb F_q((\frac{1}{T})) (x,y)$.

If we take $u$ to be any element admitting the minimal positive valuation with respect to $v$, i.e. a uniformizer of the discrete valuation ring consisting of elements of $\mathbb F_q(x,y)$ with nonnegative valuation under $v$, and $\mathbb F_{q'}$ to be the residue field of this discrete valuation ring, then the classification says $\mathbb F_{q'} ((u)) \cong \mathbb F_q((\frac{1}{T})) (x,y)$. (The only thing to check here is that $\mathbb F_{q'}$ actually lies in $\mathbb F_q((\frac{1}{T})) (x,y)$. Once an embedding it's fixed, it's straightforward to represent any element of $\mathbb F_q((\frac{1}{T})) (x,y)$ as a Laurent series in $u$ using the fact that $u$ generates the maximal ideal of the discrete valuation ring.

By the degree, I agree you mean the unique extension of the function $\deg$ whose negation is the valuation on $\Omega$. In this case, since $u$ has positive valuation, it automatically has negative degree.