Taking @VorKir's hint:
We have
$$
f(x_0) - f(x_1)
= f[x_0,x_1](x_0 - x_1)
$$
and therefore
$$
f(x_0) + f[x_0,x_1](x - x_0) - f(x_1)
= f[x_0,x_1](x - x_1).
$$
Now,
\begin{align}
f[x_0,x_1](x - x_1) - f[x_1, x_2](x - x_1)
& = (x - x_1) \left( f[x_0,x_1] - f[x_1, x_2] \right) \\
& = (x - x_1)(x_0 - x_2) f[x_0,x_1,x_2],
\end{align}
since
$$
f[x_0,x_1,x_2]
= \frac{f[x_1,x_2] - f[x_0,x_1]}{x_2 - x_0}.
$$
We have now reduced the expression down to
$$
(x - x_1)(x_0 - x_2) f[x_0,x_1,x_2] + f[x_0,x_1,x_2](x - x_0)(x - x_2) - f[x_1,x_2,x_0](x - x_1)(x - x_2).
$$
We now use that divided differences are invariant under permutations of the $x_i$ , so we have
$$
f[x_0,x_1,x_2]
= f[x_1,x_2,x_0]
$$
and therefore the expression is equal to
\begin{align}
& \ f[x_0,x_1,x_2]\left((x - x_1)(x_0 - x_2) + (x - x_0)(x - x_2) -(x - x_1)(x - x_2)\right) \\
= & \ f[x_0,x_1,x_2]\left( (x - x_1)(x_0 - x_2) + (x_1 - x_0)(x - x_2) \right) \\
= & \ f[x_0,x_1,x_2](x_0 - x) (x_2 - x_1).
\end{align}
This proof was more delicate than I expected.
First observe that by an easy induction, for all $x_0,\ldots, x_m$ we have
$$f[x_0,\ldots, x_m]=f[x_m,\ldots,x_0]\,. \quad \quad (1)$$
Recall $L_m$ is defined to be the degree $m$ polynomial interpolating $f$ at $x_0,\ldots,x_m$. Our goal is to verify by induction that
$$L_m(x)+f[x_0,\ldots,x_{m+1}] \, \omega_{m+1}(x)=L_{m+1}(x) \, \quad \quad (2) $$
for all choices of $x,x_0,\ldots, x_m.$
Since both sides of (2) are degree $m+1$ polynomials and (2) is clear for $x=x_0,\ldots,x_m$, it suffices to verify it also holds for $x=x_{m+1}$. To this end, define $Q_k$ as the degree $k$ polynomial interpolating $f$ at $x_1,\ldots,x_{k+1}$ and denote
$$\psi_m(x):=\prod_{j=1}^m (x-x_j)=\frac{\omega_{m+1}(x)}{x-x_0} \,. \quad \quad (3)$$
By the induction hypothesis, equation (2) holds for polynomials of lower degree, so
$$Q_m(x)=Q_{m-1}(x)+f[x_1,\ldots,x_{m+1}] \psi_{m}(x) \, \quad \quad (4) $$
and by considering the points in the reverse order $x_m,\ldots,x_1,x_0$
$$L_m(x)=Q_{m-1}(x)+f[x_m,\ldots,x_0] \psi_{m}(x) \, \quad \quad (5) $$
Subtracting the last two equations and setting $x=x_{m+1}$ gives
$$Q_m(x_{m+1})-L_m(x_{m+1})=\Bigl(f[x_1,\ldots, x_{m+1}]- f[x_m,\ldots,x_0] \Bigr)\psi_{m}(x_{m+1})\quad \quad \quad \quad \quad \; \; \; \;$$
$$=f[x_0,\ldots, x_{m+1}]\, \omega_{m+1}(x_{m+1}) \,, \quad \quad (6) $$
using equations (1) and (3) in the last step.
Since $Q_{m}(x_{m+1})=f(x_{m+1})=L_{m+1}(x_{m+1})$, equation (6) gives (2) for $x=x_{m+1}$. This completes the induction step and the proof of (2).
Best Answer
the answer is $ \sum_{j=0}^{n}f[x_{0}, x_{1},..., x_{j}]g[x_{j}, x_{j+1},..., x_{n}] $