Given a function $F: A \rightarrow B$ and functions $R_1, R_2, \dots, R_k:B \rightarrow A$, we can create a chain of length $k$ from a starting point $a_0$ to an end point $a_k$ using $a_i = R_i(F(a_{i-1}))$.
A rainbow table for $(F, R_1, \dots, R_k, k)$ is a collection of chains with end points $(a_0, a_k)$ organized so that searching for chains ending at $a_k$ is cheap.
We use a rainbow table to (try to) invert $F$ as follows. Given $b$, we compute $u_{11}, u_{12}, \dots, u_{1k}, u_{22}, u_{23}, \dots, u_{kk}$ using the equations $$u_{ii} = R_i(b) \qquad\text{and}\qquad u_{ij} = R_j(F(u_{i,j-1})).$$ After computing each $u_{ij}$, we check to see if there is a chain $(a_0, u_i)$ in the rainbow table. If there is, we compute $a_1, a_2, \dots, a_k$ as above and check if $F(a_j) = b$ for some $j$. If so, we have found a preimage of $b$ and we are done. Otherwise, we continue until we have checked all $u_{ij}$.
The idea is that the $R_i$ are very cheap to compute (compared to $F$), so generating a rainbow table for $(F,R_1, \dots, R_k, k)$ with $L$ entries requires storing $2L$ elements of $A$ and costs essentially $kL$ evaluations of $F$ plus the cost of organizing the table ($O(L \log L)$ or something similar?).
Each lookup costs at most roughly $k^2$ evaluations of $F$ plus the cost of $k^2$ table lookups (each $O(\log L)$ comparisons of elements of $A$ or something similar?).
The total cost for $n$ attempted inversions should be dominated by $kL + k^2n$ evaluations of $F$.
The tricky part (the part where I don't immediately know the answer) is determining fraction of elements of $A$ for which $F$ can be inverted. This is determined by the total number of distinct elements present in the chains in the rainbow table.
It would seem that if the $s$ first chains cover a fraction $\epsilon$ of $A$, then the $s+1$th chain will contain an expected $(1-\epsilon) (1-\epsilon/k)/(\epsilon/k)$ "new" elements of $A$ (or $k$ if $\epsilon/k$ is small). (The second term in the product is the expected number of iterations in the chain before it collides with a previous chain at the same index. The first term is the expected number of repetitions.)
This should mean that a rainbow table can cover a significant fraction of $A$.
When $kL$ is not too big compared to $|A|$, I would guess that the probability of inversion is close to $kL/|A|$.
If you are ok with a probability $\epsilon$ of inverting $F$ significantly smaller than $1$:
A rainbow table with parameters $k$ and $L = \epsilon |A|/k$ inverts $n \epsilon$ elements using $\epsilon |A| + k^2n$ evaluations of $F$ and storage of $2\epsilon |A|/k$ elements of $A$.
A straight-forward table for a fraction $\epsilon$ of the elements of $A$ inverts $n\epsilon$ elements using $\epsilon |A|$ evaluations of $F$ and storage of $\epsilon |A|$ elements of $A$ and $B$.
A sequence of brute force searches for a fraction $\epsilon$ of the elements of $A$ inverts $n\epsilon$ elements using $\epsilon |A|$ evaluations of $F$ and negligible storage.
Suppose you have $n=2^{10}$ target keys and have $2^{40}$ octets of fast memory available for the table. A pair of DES keys requires 14 bytes, which gives us $L \approx 2^{36}$. Choosing $\epsilon = 2^{-10}$, we find that $k=2^{-10} \cdot 2^{56} / 2^{36} = 2^{10}$, which means a total work factor of $2^{46} + 2^{30}$ DES evaluations. (If you use slow memory, memory access time will probably be much higher than DES evaluation time.)
