All you have to do is row reduce! Put everything into one matrix, and get rid of as many rows as you can.
Logically, in a basis, first of all there cannot be the zero vector. Secondly, no vector in a basis can be a linear combination of any other vectors. In your example, $X_4 = X_3 + X_2$, so you can leave out $X_4$. You are left with your basis: $X_2, X_3$
If you want to row reduce with a more complex question, you have the following matrix:
$\left[ \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 \end{array} \right]$
Move the zero row to the bottom for convenience.
$\left[ \begin{array}{cccc} 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right]$
Now subtract the first row + the second row from the third row.
$\left[ \begin{array}{cccc} 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right]$
We cannot row reduce anymore. Your answer is the rows that are not completely 0's. The first row, which maps to $X_2$, and the second row, which maps to $X_3$, is your basis. Remember, for the basis, you should use the original vectors, so the original $X_2$ and $X_3$ (coincidentally they are the same this time).
So your basis is $X_2, X_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 1\end{bmatrix},\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}$
That basis is an orthnormal basis, and therefore the coordinates of $(u,v)$ with respect to that basis are $\bigl\langle(u,v),a\bigr\rangle(=ux_1+vx_2)$ and $\bigl\langle(u,v),b\bigr\rangle(=uy_1+vy_2)$. That is$$(u,v)=(ux_1+vx_2)a+(uy_1+vy_2)b.$$
Best Answer
You ask "What exactly do we mean when say “linear” combination?"
A linear combination is an expression of the form "(scalar times object ) + (scalar times object) + ... + (scalar times object)".
Of course for such an expression to make sense you need your objects to be amenable to be multiplied by scalars and added: that's why the usual context is a vector space.
And I strongly disagree with "the term gets abused a lot": I have never seen it abused, and I cannot even imagine how it could be abused. It is among the less ambiguous terms in math.