The usual derangement formula, for permutations of $\{1,\dots,n\}$ without fixed points, is given as follows:
$$\sum_{i=0}^n (-1)^{n-i}i!\binom{n}{i} = D(n).$$
Richard Stanley, in his Enumerative Combinatorics (bottom of page 269), gives the following formula for derangements of a multiset of type $\alpha=(\alpha_1,....,\alpha_k)$:
$$D(\alpha) = \sum_{\beta_{1} =0}^{\alpha_1} ... \sum_{\beta_k=0}^{\alpha_k} \binom{\alpha_1}{\beta_1}\binom{\alpha_2}{\beta_2}...\binom{\alpha_k}{\beta_k}(-1)^{\beta_1+....+\beta_k} \binom{\sum (\alpha_i-\beta_i)}{\alpha_1-\beta_1,\alpha_2-\beta_2,...,\alpha_k-\beta_k}.$$
In the solution it says the derangement formula of $D(n)$ can be straightforwardly generalized to the sum above for the given type $\alpha=(\alpha_1,...., \alpha_k)$, where $M_\alpha$ is the multi-set $\{1^{\alpha_1},....,k^{\alpha_k}\}$.
Stanley defines a derangement of $M_\alpha$ as "a permutation $a_1a_2...a_n$ (where $\sum\alpha_i=n$) of $ M_\alpha$ such that it disagrees with every position we get by listing the elements of $M$ in a weakly increasing order, for example for the set $\{1,2^2,3\}$ the two possible derangements are $(2132,2312)$."
My question is how is that generalization straightforward? I don't see how that is achieved.
First you need to understand the formula for "simple" derangements. It's inclusion-exclusion with the following form: $$ \sum_{k = 0}^n (-1)^{k}(\text{permutations with k fixed points})(\text{number of ways to select k points to be fixed})$$ for each set of $k$ fixed points.
If there are $n - i$ fixed points, then the number of permutations is $i!$ and there are $\binom{n}{i}$ ways to select the fixed points. Note that these fixed points are not necessarily the only fixed points but they are fixed nonetheless. Inclusion-exclusion takes care of the "more fixed points than originally selected."
If you have a multiset of type $(\alpha_1,\dots,\alpha_k)$, then there are $$ \binom{\alpha_1 +\dots+\alpha_k}{\alpha_1,\dots,\alpha_k} = \frac{(\alpha_1 + \dots + \alpha_k)!}{\alpha_1! \cdots \alpha_k!} $$ multiset permutations.
Now suppose we specify $\beta_i$ of the symbol $i$ to be fixed. This gives us a total of $\beta_1 + \dots + \beta_k$ fixed points and there are
$$ \binom{\sum_i(\alpha_i - \beta_i)}{\alpha_1-\beta_1,\dots,\alpha_k - \beta_k} $$
multiset permutations of the remaining symbols. The number of ways to select these $\beta_1,\dots,\beta_k$ fixed points is
$$ \binom{\alpha_1}{\beta_1} \cdots \binom{\alpha_k}{\beta_k}. $$
So just as before, we sum over the number of ways to select the fixed points, times the number of fixed points, times $(-1)^{\text{number of fixed points}}$. This gives
$$ \sum_{\beta_1 = 0}^{\alpha_1} \cdots \sum_{\beta_k = 0}^{\alpha_k} (-1)^{\beta_1 + \dots + \beta_k} \binom{\alpha_1}{\beta_1} \cdots \binom{\alpha_k}{\beta_k} \binom{\sum_i(\alpha_i - \beta_i)}{\alpha_1-\beta_1,\dots,\alpha_k - \beta_k}. $$