The definition of linear independence says you can't make 0 out of a linear combination. It says nothing about not being able to make any other vector out of linear combinations.
(1,0) and (0,1) are independent since you cannot write (0,0) = c(1,0) + d(0,1) without c=d=0. But you can write every other vector as a nontrivial linear combination of these. (2,3) = 2(1,0)+3(0,1) for example. Spend some time making sense of the definitions with some concrete examples like this one and it will make sense eventually.
If you call your orthogonal set $\{v_1, v_2, \dots, v_n\}$, you can trivially write any vector in your set as a linear combination (take all coefficients $0$ except the coefficient of $v_k$ which is $1$).
$v_k = 0\cdot v_1+0\cdot v_2+\dots+0\cdot v_{k-1}+1\cdot v_k+0\cdot v_{k+1}+\dots+0\cdot v_n$
This is true of any set, whether it is orthogonal or not.
Moreover, any vector in the span of $\{v_1, v_2, \dots, v_n\}$ can be written as a linear combination of these vectors. This is again true of any set, whether orthogonal or not.
If you mean by formula some expression that is a continuous function of its arguments, then the answer is that this is impossible, for similar reasons to what I explained in this answer.
Suppose your $n-1$ vectors span a space of dimension $d<n-1$, then the space $S$ of possible vectors orthogonal to them has dimension $n-d>1$. Now if you take any subspace $L$ of dimension$~1$ in $S$, you can easily make that line to be the only set of possibilities by making a very small adjustment to your vectors (add small multiples of vectors in $S$ but orthogonal to $L$ to some of your vectors). By continuity, the vector of $S$ that your formula chooses must be arbitrarily close to any such line $L$, and the zero vector is the only one that satisfies this requirement.
Best Answer
Take a unit timelike vector $v$ and write $V = v^\perp \oplus \Bbb R v$. Assume that $x= x_0+av$ and $y=y_0+bv$ are lightlike and orthogonal (written according to this decomposition). Since $v$ is timelike, $v^\perp$ is a spacelike hyperplane, and so $x_0$ and $y_0$ are spacelike too. We have three relations: $$\langle x_0,x_0\rangle = a^2,\quad\langle y_0,y_0\rangle = b^2\quad\mbox{and}\quad \langle x_0,y_0\rangle = ab. $$ Then $$\langle bx_0-ay_0,bx_0-ay_0\rangle = b^2a^2 -2baab +a^2b^2=0. $$ But $bx_0-ay_0$ is spacelike, and the above then implies that $bx_0-ay_0=0$. Also note that $a\neq 0$ and $b\neq 0$, since $x$ and $y$ are lightlike. With this $$by-ax= by_0 +bav- ax_0 -abv=0, $$and so $\{x,y\}$ is linearly dependent.