Why do repunit primes have only a prime number of consecutive $1$s?

The number whose decimal representation consists of $ab$ consecutive $1$'s is divisible by the number whose decimal representation consists of $a$ consecutive $1$'s.

To see this, think of the usual "long division" algorithm. It will terminate with $0$ remainder. The quotient will have decimal representation $1$, followed by $a-1$ $0$'s, then a $1$, then $a-1$ $0$'s, and so on.

Thus if $k$ is composite, then $R_k$ is composite.

$$\frac{10^{pq}-1}{9}=\frac{10^p-1}{9}\cdot \left(10^{q(p-1)}+10^{q(p-2)}+\cdots+1\right)$$ It is easy to see that the expression on the right-hand side telescopes.

If $i$ is composite, write $i=jk$ then you have $R_i=\displaystyle\frac{10^i-1}{9}=\displaystyle\frac{10^{jk}-1}{9}=\displaystyle\frac{(10^j)^k-1}{9}=(10^{j({k-1})}+10^{j({k-2})}+...+10^j+1)\cdot \frac{10^j-1}{9}$ so if $i$ is composite then $R_i$ is composite, because of that if $R_i$ is prime then $i$ is prime.