**Warning:** *the following may not be considered as a proper answer in that it does not provide a closed form solution to the question, esp. when compared with [the previous answers][1]. I however found the approach sufficiently interesting to work out the conditional distribution.*

Consider the preliminary question of getting a sequence of $N$ heads out of $k$ throws, with probability $1-p(N,k)$. This is given by the recurrence formula
$$
p(N,k) = \begin{cases} 1 &\text{if } k<N\\
\sum_{m=1}^{N} \frac{1}{2^m}p(N,k-m) &\text{else}\\
\end{cases}
$$
Indeed, my reasoning is that no consecutive $N$ heads out of $k$ draws can be decomposed according to the first occurrence of a tail out of the first $N$ throws. Conditioning on whether this first tail occurs at the first, second, ..., $N$th draw leads to this recurrence relation.


Next, the probability of getting the *first* consecutive N heads in $m\ge N$ throws is
$$
q(N,m) =\begin{cases} 
          \dfrac{1}{2^N} &\text{if }m=N\\
    
         p(N,m-N-1) \dfrac{1}{2^{N+1}} &\text{if } N<m<2N+1
         \end{cases}
$$
The first case is self-explanatory. the second case corresponds to a tail occuring at the $m-N-1$th draw, followed by $N$ heads, and the last case prohibits $N$ consecutive heads prior to the $m-N-1$th draw. (The two last cases could be condensed into one, granted!)

Now, the probability to get $M$ heads first and the first consecutive $N$ heads in *exactly* $m\ge N$ throws (and no less) is
$$
r(M,N,m) = \begin{cases} 
          1/2^N &\text{if }m=N\\
          
         0 &\text{if } N<m\le N+M\\
          
          \dfrac{1}{2^{M}}\sum_{r=M+1}^{N}\dfrac{1}{2^{r-M}}q(N,m-r)&\text{if } N+M<m
\end{cases}
$$
Hence the conditional probability of waiting $m$ steps to get $N$ consecutive heads given the first $M$ consecutive heads is
$$
s(M,N,m) = \begin{cases} 
          1/{2^{N-M}} &\text{if }m=N\\
          0 &\text{if } N<m\le N+M\\         
\sum_{r=M+1}^{N}\dfrac{q(N,m-r)}{2^{r-M}}&\text{if } N+M<m

\end{cases}
$$
The expected number of draws can then be derived by 
$$
\mathfrak{E}(M,N)= \sum_{m=N}^\infty m\, s(M,N,m)
$$
or $\mathfrak{E}(M,N)-M$ for the number of *additional* steps...


  [1]: https://stats.stackexchange.com/a/23003/7224