Let $B$ be the graded ring $\bigoplus_i B^i$ (with $B^k B^l \subset B^{k+l}$), and $B_f$ the multiplicative group of all formal sums $1 + b_1 + b_2 + \cdots$ where $b_i \in B^i$ for all $i$.
The idea when talking about genera (such as the Todd genus or the $L$ genus) is that we can take $B$ to be $H^{2 \bullet}(X,\mathbb{Q})$ for example, and then a typical element of $B_f$ is something like the total Chern class.
Now a genus corresponds to a multiplicative sequence $(K_n)$, where each $K_n$ is a polynomial in n variables over $B$ that is homogeneous (with respect to the grading), so that $K_n(t x_1, t^2 x_2, \ldots, t^n x_n) = t^n K_n(x_1, \ldots, x_n)$. This corresponds to the idea that for an $n$ dimensional complex manifold, the evaluation of $K_n$ on the fundamental class should be some number, so that we want this homogeneity to ensure that $K_n$ corresponds to some element of the top cohomology group.
Given this, we can form an element $K(b)$ in $B_f$ for each element $b = 1 + b_1 + b_2 + \cdots$ of $B_f$ by $K(b) = 1 + K_1(b_1) + K_2(b_1,b_2) + \cdots$.
Then multiplicativity of the sequence $(K_n)$ means that $K(bc) = K(b)K(c)$, which is what you would want in algebraic topology to get multiplicativity of the numbers you get, from multiplicativity of the total Chern classes.
Now, all the books I have been reading about this say that there is an essentially unique way to associate such a multiplicative sequence to a formal power series, and that all multiplicative sequences come from this. But I don't understand precisely how; I don't even know over what ring the formal power series should be defined and even less how one can find a multiplicative sequence out of that.
It's easy enough when $B = H^{2 \bullet}(X,\mathbb{Q})$ and you're working with Chern classes, as the Chern classes of a direct sum of lines bundles are given by the symmetric polynomials in the first Chern classes. Then, given a formal power series $Q(x) \in \mathbb{Q}[[x]]$, one can form the product $\prod_i Q(x_i)$ in $\mathbb{Q}[[x_1, \ldots, x_n]]$ and get a multiplicative sequence as YBL says in his answer, by taking $1 + \sum_j K_j = \prod_i Q(x_i)$, where the $K_j$ are taken as polynomials in the elementary symmetric polynomials (ie the Chern classes).
But I don't see how you can do this without appealing to this decomposition (or indeed in a more general setting without any idea of Chern classes).
For example, in Characteristic Classes by Milnor and Stasheff, they do the same explanation as I did above, but when they get to the formal power series part it seems that they just assume $B = \Lambda[t]$ for some commutative ring $\Lambda$ and from there I lose track of what is happening; this gets me really quite confused as to what's going on.
Best Answer
I think the idea is that using the splitting principle everything reduces to the first Chern class of line bundles: Chern classes of a general bundle $E$ are symetric functions of $c_1(L_i)$ where $\bigoplus L_i = E$.
If $Q(z)$ is a power series with constant term 1, you can define $K_n$ by the formula: $$ \sum K_n(x_1,\ldots,x_n) = \prod Q(z_j) $$ where $x_i$ is the $i$-th elementary symetric function of the variables $z_j$. Hogomeneity corresponds to the fact that the $z_j$ have degree 1.
I think the statement that every multiplicative sequence comes from such a power series is only true in caracteristic 0. A multiplicative sequence with coefficients in $A$ corresponds to a ring homomorphism from the Lazard ring $\mathbb{L} = \Omega^*(pt)$ to $A$ that is to a formal group law $F(t_1,t_2) \in A[[t_1,t_2]]$. There is a natural action of power series $f(z)$ satistfying $f(0) = 0$ and $f'(0) = 1$ on the Lazard ring; just change of coordinate of formal group laws $(F^f)(t_1,t_2) := f^{-1}(F(f(t_1),f(t_2))$. Now in caracteristic zero, this action is simply transitive: every law is equivalent to the additive one $(t_1,t_2) \mapsto t_1+t_2$ because we can define the logarithm of a law by formally integrating an invariant differential. This should correspond to the fact that every multiplicative sequence is defined by a power series $Q(z) = z/f(z)$.