I thought that the seed for BBS was computed via prime multiplication.
This is not how BBS works. To help better understand the seed, it's necessary to explain BBS.
Blum Blum Shub is realized in the form of $x_{i+1} = x^2_i \bmod N$. The initial seed, $x_0$, merely has to be co-prime to Blum primes $p$ and $q$ (where $N = p\cdot q$, making $N$ a Blum integer), and must not be $0$ or $1$. The seed is $x_0$, not $p$ and $q$ (which were likely generated once to create the hardcoded $N$ and then thrown away*). They are likely using the raw hash output as the seed, since it's unlikely that it won't be co-prime to $N$. They give you the modulus on the page you linked, but not the primes:
5360751114622011236087979022735306713592537659947455285353148181
8526455500785383194936281698585494806464721601253386562653528822
5543322562532917886396992291116815877218495559211688377961466145
0857446201926232198015028137924055146650866097197909406089918491
4133568666470293429076893274969016410915119621659901793350665953
This is a 1062-bit modulus. Unfortunately, it seems that BBS can be broken† up to $10^{54} n^3$ times faster than it takes to factor the $n$-bit modulus. This means that BBS used with their parameters can be broken $10^{54} \times 1062^3 \approx 1.20 \times 10^{63}$ times faster than it takes to factor the modulus, which is a significant enough speedup that the proof is completely useless. Additionally, the proof assumes only a single bit is extracted from each $x_i$ (officially by taking the parity of $x_i$). If more bits are extracted at a time, it becomes even weaker. Parameters that seem acceptable are terrible for BBS.
The security proof for BBS is only a reduction to a problem with no known solution. That is, a correct implementation (i.e. one where only one bit is extracted from each $x_i$, and only when the modulus is "sufficiently" large) can only be broken if the quadratic residuosity problem (QRP) is solved.‡ It is "proven secure" in the same way that RSA is "proven secure" because a proper implementation is provably reducible to the security of the RSA problem (the difficulty of finding the $e^\mathrm{th}$ root of an integer, modulo semiprime $N$), but no one thinks of RSA as having perfect security. In other words, BBS has a security reduction to a yet-unproven problem, and is not information-theoretic secure.
How can above steps 600 and 605 be connected together?
Are you asking how the 140-byte seed comes from the 160-bit hash? That is actually answered in the link you provided. There are seven SHA-1s. The seventh hashes every seventh byte, the sixth hashes every sixth byte, etc. The bytes come from an image of the lava lamp itself. The final hash values are concatenated to form a 140-byte (1120-bit) value. This value is used, directly, as $x_0$.
* The only thing you are losing by not having access to $p$ and $q$ is the ability to calculate any $x_i$ value directly from $x_0$ without having to calculate $x_1 \cdots x_{i-1}$ first. Euler's theorem can be used to do $x_i = (x_0^{2^i \bmod \lambda(N)}) \bmod N$ where $\lambda$ is the Carmichael totient function. If an attacker factors $N$, they can do this as well, breaking the security reduction.
† Broken in this context does not refer to the ability to predict the random stream or recover the seed. Instead, it means the security proof is invalidated and that the algorithm is no longer reducible to a known hard problem and that a break could exist which does not rely on solving QRP. Critically, it does not automatically imply that such a break does, in fact, exist.
‡ Although the quadratic residuosity problem is often thought to be as difficult as integer factorization, this has not been proven. It is possible that a method to break BBS could be exist which solves QRP without solving integer factorization.
