Consider $F$ be twice differentiable function with continuous derivatives. Let $X$ be semimartingale, i.e. $X_t = X_0 + M_t + A_t$, where $M_t$ is continuous local martingale and $A_t$ is continuous process with bounded variation.
We want to show that: $F(X_t) = F(X_0) + F'(X_t)\cdot X_t + \dfrac{1}{2} F''(X_t) \cdot \langle M_t\rangle$.
The proof based on: $F(X_t) - F(X_0) = \displaystyle \sum_{i=0}^{n} F'(X_{t_{i-1}}) (X_{t_i} - X_{t_{i-1}}) + \frac{1}{2} \sum_{i=0}^{n} F''(\eta_i)(X_{t_i} - X_{t_{i-1}})^2 = S_1 + S_2$.
Now we need to prove that $S_1 \to F'(X)\cdot X$ and $S_2 \to \dfrac{1}{2} F''(X) \cdot \langle M_t\rangle$.
First part is quite obvious(just use Lagrange-formula for $F'(X_{t_{i}}) - F'(X_{t_{i-1}}))$.
But the second part is quite unclear. My professor writes this (all limits are in probability): $$\lim_{n}\sum_{i=0}^{n}F''(X_{t_{i-1}})(X_{t_{i}} - X_{t_{i-1}})^2 = \lim_{n}\sum_{i=0}^{n}F''(X_{t_{i-1}})(M_{t_{i}} - M_{t_{i-1}})^2 = \dots.$$
I don't understand where is the $A_{t_{i}} - A_{t_{i-1}}$ term. Obviously we may use $X_t = X_0 + M_t + A_t$, but how $A_t$ vanishes?
The $(A_{t_i}-A_{t_{i-1}})^2$ terms go away because $A$ has bounded variation. Let $V(t)$ be the variation of $A$ at time $t$ and assume $F'' \le K$ by localizing $X$ and using that a continuous function is bounded on a compact domain. Then
\begin{align*} \lim_n \sum_{i=0}^n F''(X_{t_{i-1}})(A_{t_i}-A_{t_{i-1}})^2 &\le \lim_n K\sum_{i=0}^n (A_{t_i}-A_{t_{i-1}})^2 \\ &\le K \lim_n \left( \sup_{i \le n} |A_{t_i}-A_{t_{i-1}}| \right)\sum_{i=0}^n |A_{t_i}-A_{t_{i-1}}| \\ &\le K V(t) \lim_n \left( \sup_{i \le n} |A_{t_i}-A_{t_{i-1}}| \right) \\ &= 0 \end{align*}
by the continuity of $A_t$. A similar argument shows the $(A_{t_i}-A_{t_{i-1}})(M_{t_i}-M_{t_{i-1}})$ terms go to $0$ in the limit.