Few questions regarding this proof:
1) As I understand the reason for the third line is to make the order of derivative for each function equal. The bit I don't understand is why k=1 under the sum symbol? Why after the sum symbol $k$ changes to $k-1$?
2) On the fourth line highlighted n has changed from $n+1$ because we need to make the sums equal? And if it had stayed the same how the last member highlighted in yellow would look like ( I know that it wouldn't be convenient and useful, just want to understand the reasoning).
3) Why does derivatives of k=0,n disappear on the last line?
Step $3$ is called reindexing the sum. Instead of having the dummy variable $k$ range over the values from $0$ to $n$, you have it range over the values $1$ to $n+1$, and modify the quantity being summed to compensate for the shift. In general, shifting a summation by $r$ places looks like $$ \sum_{k=a}^b c_k=\sum_{k=a-r}^{b-r}c_{k+r} $$ Take the time to convince yourself this equation is true, by expanding out both sums and seeing they are equal to $c_a+c_{a+1}+c_{a+2}+\dots+c_b$.
In step $4$, they "pull out" the $k=0$ term $\binom{n}0f^{(n+1)}g$ from first summation and put it out front, leaving the smaller summation $\sum_{k=1}^n f^{(n+1-k)}g^{(k)}$. Similarly, they remove the $k=n$ term from the other summation and put it on the right, leaving a smaller summation. In general, this looks like \begin{align} \sum_{k=a}^n c_k &=c_a+\sum_{k=a+1}^b c_k\tag{pulling out from front}\\ &=\left(\sum_{k=a}^{b-1}c_k\right)+c_b\tag{pulling out from back} \end{align}
In step $7$, the terms at the ends, which were pulled out of the summation before, are pulled back into the summation. Notice how the summation in the previous line goes from $k=1$ to $n-1$, and in the last line goes from $k=0$ to $n$. The two new terms, $k=0$ and $k=n$, in the summation are precisely the terms that "disappeared." To see this is true, start with the last expression, and see what happens when you "pull out" the first and last term.