Here is the proof, I understand it for the most part, but have some specific questions to help me understand better.
In the induction step:
Line 2, we multiply $(1+x)$ to balance the comparison so we can utilize the induction hypothesis?
Line 4, how are we able to add $(1+x)$ to balance the inequality, and then remove $kx^2$ freely? How do I know when I can add 'stuff' and remove 'stuff'?
On
The trick is to understand each equality and inequality separately, and then you can chain them together.
Line 1: $(1+x)^{k+1} = (1+x)^{k} (1+x)$ - this is one of the properties of powers.
Line 2: We assume that our inequality is true for $k$: $(1+x)^{k} \geq 1+kx$ - this is our induction hypothesis. The equality remains true if we multiply both sides by a positive value. $1+x > 0$ because $x>-1$. Multiplying the hypothesis inequality by $1+x$ yields: $(1+x)^{k} (1+x) \geq (1+kx) (1+x)$
Line 3: algebra
Line 4: $kx^2 \geq 0$ (because $k>0$). If we add $1+(k+1)x$ to both sides, the inequality will remain true: $1+(k+1)x + kx^2 \geq 1+(k+1)x$
The trick is that we need to prove a result about $(1+x)^{k+1}$, but all we have is information about $(1+x)^{k}$. Thus, in line one have to manipulate $(1+x)^{k+1}$ in a way that lets us use what we know about $(1+x)^{k}$ In particular, for that $k$ we do know that $(1+x)^k \geq 1+kx$ by the induction hypothesis. Now it follows that $$(1+x)(1+x)^k \geq (1+x)( 1+kx)$$ because $(1+x)$ is a positive quantity, so it won't change the direction of the inequality if we multiply it to each side. Now we can do some algebra, as done in line three above.
In line four, the author completely drops the $kx^2$ term. This is a great move for two reasons: First, because you are guaranteed that the $kx^2$ terms is not negative. Thus, by removing it you know that the RHS will either become smaller or stay the same. Second, by removing the $kx^2$ term, you end up with the EXACT inequality that you want to end the proof by induction. Note that if you plug $k+1$ for $n$ into the inequality $(1+x)^n \geq 1+nx$, you will be exactly at the point where $kx^2$ is dropped. This is tantamount to showing that the inequality holds for the $(k+1)^{th}$ case, so you are done.