There are two things going on here. First, $\displaystyle \frac{4!}{2^{2}6^{4}}$ is about $0.0046$, not $0.46$.
Second, the problem described by Feller and your simplification to four dice are not the same problem. Feller's problem requires that each face appear twice. Your four-dice problem requires that $some$ two faces appear twice. There's only one way that Feller's event can be satisfied; namely, that each of 1, 2, 3, 4, 5, and 6 appear exactly twice. There are many more ways that some two faces could appear twice. For instance, you could have two 1's and two 2's, two 1's and two 3's, etc. In fact, there are $\binom{6}{2} = 15$ different ways to choose the two faces that will appear.
And that solves your problem. Taking the correct result from Feller, $\displaystyle \frac{4!}{2^{2}6^{4}} \approx 0.0046$, and multiplying it by the $15$ different ways to choose the faces gives $\displaystyle \frac{90}{6^4} \approx 0.0694444$.
With respect to your question about the multinomial coefficient, $\displaystyle \binom{n}{k_1, k_2, \ldots, k_m}$ calculates the number of ways to partition a set of $n$ items into $m$ groups with $k_1$ items in the first group, $k_2$ items in the second group, and so forth. So in Feller's problem $\displaystyle \binom{12}{2, 2, 2, 2, 2, 2}$ is calculating the number of ways to put two dice in the 1's group, two dice in the 2's group, and so forth.
In the four-dice problem you describe, you haven't specified a particular partition of dice into groups; there are several partitions that will satisfy "two faces appearing twice." So, in the four-dice problem, $\displaystyle \binom{4}{2, 2}$ counts the number of ways to put two dice in a 1's group and two dice in a 2's group. It also counts the number of ways to put two dice in a 1's group and two dice in a 3's group, and it counts the number of ways to put two dice in a 1's group and two dice in a 4's group, and so forth. So to calculate the total number of ways of obtaining "two faces appearing twice" you need to multiply $\displaystyle \binom{4}{2, 2}$ by the number of ways to pick two of the faces out of six possible faces, which is $\binom{6}{2}$. Then you mutiply by $\frac{1}{6^4}$ to get the probability you want.
If you want a general answer, the probability of throwing $n$ $d$-sided dice and having exactly $m$ faces appear $\frac{n}{m}$ times would be
$$\binom{n}{m} \frac{n!}{((\frac{n}{m})!)^m d^n} .$$
Using the notation $A^{k}$ to mean $k$ $A$'s:
$$\begin{eqnarray}
|A^{1+} \cap B^{2+}| &=& |A^{1+}| + |B^{2+}| - |A^{1+}\cup B^{2+}| \\
&=& 3^n - |A^0| + 3^n - |B^0| - |B^1| - (3^n - |A^0\cap B^0| - |A^0\cap B^1|) \\
&=& 3^n - 2^n + 3^n - 2^n - n2^{n-1} - (3^n - 1 - n) \\
&=& 3^n - 2^{n+1} - n2^{n-1} + n + 1
\end{eqnarray}$$
So your probability is:
$$P(n) = \frac{3^n - 2^{n+1} - n2^{n-1} + n + 1}{3^n}$$
Best Answer
Let $a_i, b_i\in[0,1]$ be the probabilities that die $A, B$ shows $i+1$. (According to the edit, $i$ can only range over $\{0,1,2,3,4,5\}$ for both dice). Let $f(z)=\sum a_iz^i$ and $g(z)=\sum b_i z^i$. Then the desired result is that $$f(z)g(z)=\sum_{j=0}^{10}\frac1{11}z^j,$$ a polynomial without real roots. Hence both $f$ and $g$ have no real roots, i.e. they must both have even degree. This implies $a_5=b_5=0$ and the the coefficient of $z^{10}$ in their product is also $0$, contradiction.
Remark: Why does $h(z)=\sum_{j=0}^{10}z^j$ not have real roots? We have $h(z)=\frac{1-z^{11}}{1-z}$ and the numerator has only one real root at $z=1$. But that is no root of $h$.