The proof boils down to integrating twice:

$$\begin{aligned}\int_a^b(x-a)(b-x)f''(x)\,\dd x &= \bigl[(x-a)(b-x)f'(x)\bigr]_a^b-\int_a^b(a+b-2x)f'(x)\,\dd x\\ \\ &=\int_a^b(2x-a-b)f'(x)\,\dd x\\ \\ &=\bigl[(2x-a-b)f(x)\bigr]_a^b-\int_a^b 2f(x)\,\dd x\\ \\ &= -2\int_a^b f(x)\,\dd x\end{aligned}$$

The tags of the partition are $x_k=a+kh$ for $k=0,1,\ldots n$ with $h=\frac{b-a}{n}$ and the trapezoidal rule is given by $$T=\frac{h}{2}\bigl(f(a)+f(b)\bigr)+h\sum_{k=1}^{n-1} f(a+kh)$$ Firt we consider the sub-interval $[x_{k-1},x_{k}]$ for $k=1,\ldots n$ and estimate the contribution to the truncation error on each sub-interval. We introduce the function $$g(x)=f(x)-f(x_{k-1})-\frac{\bigl(f(x_{k})-f(x_{k-1})\bigr)\bigl(x-x_{k-1}\bigr)}{h}$$ The function $g$ is the deviation of $f$ on the interval $[x_{k-1},x_{k}]$ with the linear function between the points $\bigl(x_{k-1},f(x_{k-1})\bigr)$ and $\bigl(x_{k},f(x_{k})\bigr)$. Then $\displaystyle \int_{x_{k-1}}^{x_k}g(x)\,\dd x$ is equal to the contribution of this sub-interval to the truncation error of the trapezoidal rule. Because the function $g$ is twice differentiable, and by definition is $g(x_{k-1})=g(x_{k})=0$, the lemma is applicable: $$\int_{x_{k-1}}^{x_k}g(x)\,\dd x=-\frac{1}{2}\int_{x_{k-1}}^{x_k} (x-x_{k-1})(x_{k}-x)g''(x)\,\dd x$$ By definition, $g''(x)=f''(x)$ is valid and therefore: $$\begin{aligned}\left|\int_{x_{k-1}}^{x_k}g(x)\,\dd x\right| &\le \frac{1}{2}\int_{x_{k-1}}^{x_k} (x-x_{k-1})(x_{k}-x)|f''(x)|\,\dd x \\ \\ &\le \frac{M}{2} \int_{x_{k-1}}^{x_k} (x-x_{k-1})(x_k-x)\,\dd x \\ \\ &=\frac{M}{2}\int_{x_{k-1}}^{x_k} (-x^2+(x_{k-1}+x_k)x-x_{k-1}x_{k})\,\dd x\\ \\ &= \frac{M}{12}(x_{k}-x_{k-1})^3\\ \\ &= \frac{M}{12}h^3\end{aligned}$$ Then we have for the truncation error of the trapezoidal rule we have $$\left|\int_a^bf(x)\,\dd x - T\right|\le\sum_{k=1}^n \tfrac{1}{12}Mh^3=\tfrac{1}{12}Mh^3n=\tfrac{1}{12}M(b-a)h^2$$