Say, $g: X\to Y$ and $f: Y\to Z$ are smooth.
One can find $(f \circ g)'''$ by using the Faà di Bruno's formula:
$$(f \circ g)''' =(f'''\circ g)(g')^3 + 3(f''\circ g)g'g'' + (f'\circ g)g'''$$
But my current level doesn't allow me to understand $(g')^3$ (I guess it has something to do with tensors yet they are well beyond my text).
So I wanted to express it in more familiar terms by recalling that
$$(f \circ g)''':X\to \mathcal L^3(X;Z)$$
would it be something like
\begin{align*} (f\circ g)'''(x)[a,b,c] & = f'''(g(x))\Big[g'(x)[a], g'(x)[b],g'(x)[c]\Big] \\ & + 3f''(g(x))\Big[g'(x)[a],g''(x)[b,c]\Big] \\ & + f'(g(x))\Big[g'''(x)[a,b,c]\Big] \end{align*}
Does it make sense? How can we derive the extended version per hand?