Let $(X_t)_{t\geq 0}$ be an $(\mathcal{F}_t)_t$- Brownian Motion and let $f\in C_b^2(\mathbb{R}^d)$. For $s\leq t$, by Taylor's theorem, $f(X_t)=f(X_s)+(X_t-X_s)f'(X_s)+\frac{1}{2}(X_t-X_s)^2f''(X_s)+(X_t-X_s)^2E_1(s,t)$, where $E_1(s,t)=\int_0^1(1-u)(f''(uX_t+(1-u)X_s)-f''(X_s))du$.
My question: how can we exactly write the error term in this way?
Define the function $g: [0,1] \to \mathbb{R}$ by $g(u) = f(u X_t + (1-u) X_s) = f(X_s + u(X_t - X_s))$. Now apply Taylor's Theorem with the integral form of the remainder to $g$ to obtain $$g(1) - g(0) - g'(0) = \int_0^1 (1 - u) g''(u) du.$$
Computing all the derivatives, this is the same as $$f(X_t) = f(X_s) + (X_t - X_s) f'(X_s) + (X_t - X_s)^2 \int_0^1 (1-u) f''(uX_t + (1-u) X_s) du.$$
This is the same as your desired formula since $$(X_t - X_s)^2 \int_0^1 (1-u) f''(X_s) du = \frac{1}{2} (X_t-X_s)^2 f''(X_s).$$