We are given two plaintext/ciphertext pairs $(P_1, C_1)$ and $(P_2, C_2)$ with $E_{k_1^*}(P_i) = D_{k_2^*}(C_i)$ where the true key is $k^* = (k_1^*, k_2^*)$. Suppose the cipher key is $n$ bits long, and the block is $b$ bits long. (Schneier uses $m$ for the block size, but as a confusion of minims I cannot keep these letters straight, so I will use the much more sensible letter $b$ for block size.)
Let's model the $k_i^*$ as independent uniform random bit strings, because that's how a sensible user will choose them. For simplicity, let's model $E_{k_1}$ and $D_{k_2}$ as independent uniform random functions. Obviously this model is wrong because they're both permutations, and if $k_1 = k_2$ then they can't be independent, but $\Pr[k_1 = k_2] = 1/2^n$ so this event happens only a negligible fraction of the time, and all models are wrong—some are just useful.
Suppose we find a candidate key $k = (k_1, k_2)$ such that $E_{k_1}(P_1) = D_{k_2}(C_1)$. What is the probability that $k = k^*$ given this—that is, what is the probability that we have found the true key given that we have found a matching candidate?
A priori, $\Pr[k = k^*] = 2^{-2n}$ because the true key $k^*$ is uniformly distributed among $2^{2n}$ possibilities. And obviously if $k = k^*$, then $E_{k_1}(P_1) = D_{k_2}(C_1)$ with probability $1$; whereas if $k \ne k^*$, then in the independent uniform random function model for $E$ and $D$, we have $E_{k_1}(P_1) = D_{k_2}(C_1)$ with probability $2^{-b}$. The unconditional probability that $E_{k_1}(P_1) = D_{k_2}(C_1)$ is
\begin{align*}
\Pr&\bigl[E_{k_1}(P_1) = D_{k_2}(C_2)\bigr] \\
&= \Pr[E_{k_1}(P_1) = D_{k_2}(C_2) \bigm| k = k^*\bigr] \Pr[k = k^*] \\
&\qquad + \Pr[E_{k_1}(P_1) = D_{k_2}(C_2) \bigm| k \ne k^*\bigr] \Pr[k \ne k^*] \\
&= 2^{-2n} + 2^{-b} (1 - 2^{-2n}).
\end{align*}
So by Bayes' rule, the conditional probability that the key is correct given that it matches a known plaintext/ciphertext pair is
\begin{align*}
\Pr&\bigl[k = k^* \bigm| E_{k_1}(P_1) = D_{k_2}(C_1)\bigr] \\
&= \Pr[k = k^*]
\frac{\Pr\bigl[E_{k_1}(P_1) = D_{k_2}(C_2) \bigm| k = k^*\bigr]}
{\Pr\bigl[E_{k_1}(P_1) = D_{k_2}(C_2)\bigr]} \\
&= 2^{-2n}
\frac{1}
{2^{-2n} + 2^{-b}(1 - 2^{-2n})} \\
&= \frac{1}{1 + 2^{-b} (2^{2n} - 1)} \\
&= \frac{2^b}{2^b + 2^{2n} - 1}.
\end{align*}
For $b \ll 2n$, this is roughly $2^{b - 2n}$. For $b = 2n$, this is roughly $1/2$. For $b \gg 2n$, this is roughly $1$.
What if we also confirm $E_{k_1}(P_2) = D_{k_2}(C_2)$? Then it is roughly as if the block size were $2b$ instead of $b$ in the above analysis since we are considering whether $E'_{k_1}(P_1, P_2) = D'_{k_2}(C_1, C_2)$ where $E'_{k_1}(P_1, P_2) = \bigl(E_{k_1}(P_1), E_{k_1}(P_2)\bigr)$ and likewise for $D'$. In this case, the conditional probability of a true positive is about $2^{2b - 2n}$, or $2^{3b - 2n}$ with three blocks, and so on, until the total size of the blocks reaches the size of the combined key at which point the probability rapidly converges to $1$.
In conclusion, I think you are correct that Schneier mixed up the sense of the ‘success probability’ (better named ‘true positive probability’): it should have just been $2^{2m - 2n}$, $2^{3m - 2n}$, etc., not $1$ in $2^{2m - 2n}$, $1$ in $2^{3m - 2n}$, etc., where $m$ is the confusingly named block size.
It is a small error in a book that is otherwise full of grievous intellectual harm.