Applying the induction hypothesis indirectly? I have trouble understanding this proof.

Yes, the statement $P(n)$ that you prove is

For every $n+1$ times differentiable function $R$, ...

and has only $n$ as free variable. Everything else, in particular $R$ is quantified within $P(n)$. For induction, you show $P(0)$, i.e.,

For every (once) differentiable function $R$, ...

and you show $\forall n\colon P(n)\to P(n+1)$ to finally arrive at

For all $n\in\Bbb N_0$, for every $n+1$ times differentiable function $R$ ...

For the induction step $\forall n\colon P(n)\to P(n+1)$, we proceed as follows: Let $n\in\Bbb N_0$ be arbitrary. We want to show $P(n)\to P(n+1)$. So assume $P(n)$. Now we want to show $P(n+1)$. So let $R$ be an arbitrary $n+2$ times differentiable with $R(a)=\ldots=R^{(n+1)}(a)=0$. Then $R'$ is $n+1$ times differentiable with $(R')'(a)=\ldots=(R')^{(n)}(a)=0$, hence $P(n)$ applies to $R'$ ... ... hence the conclusion holds for $R$. As $R$ was arbitrary, we conclude that $P(n+1)$ holds. Thus we have shows $P(n)\to P(n+1)$. As $n$ was arbitrary, we conclude $\forall n\in\Bbb N_0\colon P(n)\to P(n+1)$.

We have taken the function $R $ such that it is $n+2$ times differentiable and $R^{(k)}(a)=0$,$k=0,1,\ldots,n+1$. Our Induction assumption is that the proposition holds for for all functions with degree less than that of $R$. Now if we take $R'$, it has one degree less than that of $R$, it is $(n+1) $ times differentiable and since $R^{(k)}(a)=0$, $k=0,1,\ldots,n+1$, we have $(R')^{(k)}(a)=0$, $k=0,1,\ldots,n$. Hence we can apply the induction assumption for $R'$ and proceed as shown in the textbook.