Suppose that you have $k$ parallel lines. Then we might as well consider $R_n$ only for $n\ge k$. Clearly $R_k=k+1$. Suppose that you have $n$ lines for some $n\ge k$, and you add a new line that is not parallel to any of the $n$ existing lines. It will cross each of those lines. In order to cross $n$ lines, it must pass through $n+1$ regions. (Why?) It cuts each of those $n$ regions in two, so it adds $n+1$ new regions. Thus, the recurrence is the same as in the case of no parallel lines: $R_{n+1}=R_n+n+1$, or, if you prefer, $R_n=R_{n-1}+n$. Only the initial value has changed.
Added: Specifically, you have
$\begin{align*} &R_k=k+1\\ &R_n=R_{n-1}+n\quad\text{if }n>k\;. \end{align*}$
You didn’t ask about getting a closed form for $R_n$, but that’s not too hard to do by very elementary techniques. Look at the first few values:
$\begin{align*} R_k&=k+1\\ R_{k+1}=R_k+(k+1)&=(k+1)+(k+1)\\ R_{k+2}=R_{k+1}+(k+2)&=(k+1)+(k+1)+(k+2)\\ R_{k+3}=R_{k+2}+(k+3)&=(k+1)+(k+1)+(k+2)+(k+3)\;. \end{align*}$
At this point it’s not hard to guess the general formula:
$\begin{align*} R_{k+m}&=(k+1)+\sum_{i=1}^m(k+i)\\ &=k+1+\sum_{i=1}^mk+\sum_{i=1}^mi\\ &=k+1+km+\frac{m(m+1)}2\\ &=1+k(m+1)+\frac{m(m+1)}2\\ &=1+\frac12(m+2k)(m+1)\;. \end{align*}$
Set $n=k+m$, and this becomes
$R_n=1+\frac12(n+k)(n-k+1)$
for $n\ge k$.
To prove that the closed form is actually correct you would use induction on $n$ with base case $n=k$.