Referring to those two lines, can someone please explain how those results were obtained?
My understanding is, the following formula is being referenced:
$$dV_t = dV(S_t,t) = \frac{\partial V}{\partial x}S_t + \frac{\partial V}{\partial t}dt+\frac{1}{2}\frac{\partial^2 V}{\partial x^2}(dS_t)^2$$
But I am not sure how to apply it, to obtain the result:
$$dV_t = 2W_tdW$$
For instance, how do we get a $-dt$ term and a $\frac{1}{2} * 2dt$ and a $2W_tdW$ term?
We have that $$ V = W_t^2 - t$$ hence (written with $t$ and $x$, where we plug in $S_t = W_t$ for $x$), we have $$ V(x,t) = x^2 - t$$ Now \begin{align*} \frac{\partial V}{\partial t} &= -1\\ \frac{\partial V}{\partial x} &= 2x\\ \frac{\partial^2 V}{\partial x^2} &= 2 \end{align*} Hence (note that $S_t = W_t$) \begin{align*} \frac{\partial V}{\partial t}(W_t, t) &= -1\\ \frac{\partial V}{\partial x}(W_t, t) &= 2W_t\\ \frac{\partial^2 V}{\partial x^2}(W_t, t) &= 2 \end{align*}
Ito gives $$\tag{$*$} dV = 2W_t dW_t - 1 \cdot dt + \frac 12 \cdot 2 \cdot (dW_t)^2 $$ (note that you have a typo in your Ito, it should read $$ dV = \frac{\partial V}{\partial x} \color{red}{d}S_t + \frac{\partial V}{\partial t} dt + \frac 12\frac{\partial^2V}{\partial x^2}(dS_t)^2 $$ Continuing with $(*)$, as $(dW_t)^2 = dt$, we end up with $$ dV = 2W_tdW_t - dt + dt = 2W_tdW_t. $$