Let $v_1,v_2$ be vector fields on $\mathbb{R}^n$ and $\phi\in C^\infty(\mathbb{R}^n)$, is there a way to show there exists a unique vector field $w$ such that $L_w \phi=L_{v_1}(L_{v_2}\phi)-L_{v_2}(L_{v_1}\phi)$ without explicit computation ?
Aside question: what does Lie derivative tell you, what is its interpretation, when is it used?
By the way, is the Lie derivative generally vector valued or real valued?
You can take the Lie derivative of any tensor field along a vector field. Let's first deal with the case you've mentioned, where $\phi \in C^\infty(\mathbb{R}^n)$ i.e. $\phi$ is a function. For functions, the Lie derivative is pretty much the same as the directional derivative. How so?
Suppose you have a vector field $v$. Then given $\phi$, at any point $p$, you can take the directional derivative $D_v(\phi)(p) = (\nabla \phi \cdot v_p) (p)$. Thus, you can define a function $\mathcal{L}_v \phi (p) = (\nabla \phi \cdot v_p) (p)$ i.e. pointwise take the directional derivative and evaluate.
Given this information, see if you can find a vector field that would satisfy this.
In general, the Lie derivative tells you the change in a tensor field along the flow of the vector field. Tensor fields are, well, tensorial, so they have transformation rules. Using these transformation rules, you can essentially calculate this change by "pulling" and "pushing" the tensor fields along the flows. This is all hand-wavey, and can be made more explicit if you're interested.
But as a result, the Lie derivative returns the same kind of object it acts upon. If it acts on a function, it spits a function out. If it acts on another vector field, it spits a vector field out, and so on and so forth.
Hints for the actual problem on hand: Uniqueness is straightforward! First note that (for functions) you have $\mathcal{L}_v - \mathcal{L}_w = \mathcal{L}_{v-w}$ (use the gradient/dot product formulation). Now you have that $\mathcal{L}_{v-w}\phi = 0$ for all $\phi$. What does this tell you? Existence is also not bad. If you show that the right side is a derivation (i.e. satisfies the Liebniz rule), then it has to come from a directional derivative! If this is not a fact you know, take $\phi$ to be each of the coordinate functions. Note that for any coordinate function $x_i$, we have that $\mathcal{L}_v (x_i)$ spits out the $i$'th coordinate of $v$. So, putting in all the $x_i$ gives you a formula for each coordinate of $w$. This is essentially the proof of showing a derivation comes from a directional derivative!