I'm going to call your "streams" "strings" instead, because "streams" looks too much like "streaks" to me. This becomes a much easier problem if we translate it into an enumeration problem. Since each of the 210 possible bit strings occur with equal probability, it suffices to count the total number of streaks of length k in all these strings (and then as Kristal says just divide by 210 for the expected value per string). Let S(k,n) be the number of k-streaks in all strings of length n. For k = 1 we can make a straightforward recursion:
$S(1,n) = 2S(1,n-1) + 2^{n-1} - 2^{n-2} = 2S(1,n-1) + 2^{n-2}$
since for every n string, we have two copies of every (n-1)-string as a prefix (with their associated 1-streaks), plus half the time we add a new 1-streak at the end (if the bit we add at the end is different from the old last bit). Sometimes, though, we break a 1-streak at the end; this happens if our previous last bit was a 1-stream, so it happens once for each possible n-2 string as a prefix. (Note that this means the recursion requires n > 2!).
This recursion is easy to solve; we have S(1,2) = 4 from which it follows $S(1,n) = (n+2)2^{n-2}$.
Now there's a bijection between (k-1)-streaks in (n-1)-strings and k-streaks in n-strings (just add / remove another bit to the streak), so S(k-1,n-1) = S(k,n). We conclude
$S(k,n) = (n-k+3)2^{n-k-1}$
for k < n, and S(n,n) = 2. For n=10, we have the sequence
3072, 1408, 640, 288, 128, 56, 24, 10, 4, 2
for streaks of length 1, 2, 3, ..., 10 out of the 210 = 1024 possible strings, so dividing through should give the expected number in a random string.
The sequence S(1,n) is not in Sloane, but the total number of streaks in all n-strings
$\sum_{k=1}^n S(k,n)$
is - this sequence starts 2, 6, 16, 40, 96, 224,... and has general form
$\sum_{k=1}^n (n-k+3)2^{n-k-1} = (n+1)2^{n-1}$.
This sequence A057711 apparently has quite a few combinatorial interpretations.