The statement is indeed true.
Let $h(x)=\frac{2(f(x)-g(x))}{(x-a)(x-b)}$. By the intermediate value theorem, it suffices to show $m \leq h(x) \leq M$, where $m$ and $M$ are the minimum and maximum of $f''$ respectively on $[a,b]$. Rescaling if necessary, we may assume that $a=0$ and $b=1$.
Now define a map $\phi : [0,1] \to {\mathbb R}$ by
$ \phi(t)=\left\lbrace \begin{array}{lcl} (1-x)t, &\text{if} & t \in [0,x] \\ (1-t)x, &\text{if} & t \in [x,1] \end{array}\right. $
Then $\phi$ is continuous, positive, and the integral of $\phi$ equals $\frac{x(1-x)}{2}$. And
$ \begin{array}{lcl} \int_{0}^{x} \phi(t)f''(t) dt &=& (1-x)\int_{0}^{x} tf''(t) dt=(1-x)\big[tf'(t)-f(t)\big]^{x}_{0}= (1-x)\big(xf'(x)-f(x)+f(0)\big) \\ \int_{x}^{1} \phi(t)f''(t) dt &=& x\int_{x}^{1} (1-t)f''(t) dt=x\big[(1-t)f'(t)+f(t)\big]^{1}_{x}= x\big(-(1-x)f'(x)-f(x)+f(1)\big) \end{array} $
We deduce
$ \int_{0}^{1} \phi(t)f''(t) dt=(1-x)f(0)-f(x)+xf(1)=\frac{x(1-x)}{2}h(x) $
Since $-m \leq f'' \leq M$, the left-hand side is between $m\frac{x(1-x)}{2}$ and $M\frac{x(1-x)}{2}$. So $m \leq h(x) \leq M$ as wished.