Options on Futures Black-Scholes

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I am taking the Financial Risk Management course, and the topic now is "Variations on the Black-Scholes Model". I am following Paul Wilmott's "The Mathematics of Financial Derivatives: A Student Introduction."

I am currently trying to through "Options on Futures" part and having trouble following the author:

Here is Black-Scholes equation he uses:

$\frac{\delta V}{\delta t} + \frac{1}{2} \sigma^2 S^2 \frac{\delta^2 V}{\delta S^2} + rS\frac{\delta V}{\delta S} -rV = 0$

and derivation for Option on Futures:

" Option on futures have a value that depends on F and t, i.e. of the form $V(F,t)$. Since $F=Se^{r(T-t)}$, we can derive a partial differential equation for $V(F,t)$ from the ordinary Black-Scholes equation via the change of variable rule. That is, we replace $S$ by $Fe^{r(T-t)}$ throughout, and we replace

$\frac{\delta}{\delta S}$ by $\frac{\delta F}{\delta S} \frac{\delta}{\delta F} = e^{r(T-t)} \frac{\delta}{\delta F}$

and

$\frac{\delta}{\delta t}$ by $\frac{\delta}{\delta t} + \frac{\delta F}{\delta t} \frac{\delta}{\delta F} = \frac{\delta}{\delta t} - rF\frac{\delta}{\delta F}$

The result is

$\frac{\delta V}{\delta t} + \frac{1}{2} \sigma^2 \frac{\delta^2 V}{\delta F^2} - rV = 0$ "

I understand the replacement: $\frac{\delta}{\delta S}$ by $\frac{\delta F}{\delta S} \frac{\delta}{\delta F}$.

Here are my questions:

How do we justify this replacement?: $\frac{\delta}{\delta t}$ by $\frac{\delta}{\delta t} + \frac{\delta F}{\delta t} \frac{\delta}{\delta F}$

How do we get from $\frac{1}{2} \sigma^2 S^2 \frac{\delta^2 V}{\delta S^2}$ to $\frac{1}{2} \sigma^2 \frac{\delta^2 V}{\delta F^2}$ ? Aren't we missing $F^2$?

Thank you!