This is a completely different algorithm, which I believe works in $O(m\log m)$ worst case, and should work for integer or real numbers.
Let us assume that $A$ and $B$ are already in ascending order, otherwise spend $O(n\log n+ m\log m)$ to sort them. We slightly strengthen the requirement for algorithm $\mathcal{A}(A, B)$ to return all indices $i$ such that $A$ can be mapped to $B$ with offset $k=b_i-a_1$, meaning that the mapping starts at $b_i$ onwards. The high-level idea is to solve the sub-problems corresponding to a subarray of $A$ and merge the indices in a way that only valid solutions remain.
The recursion, however, depends on how close $A$ is to an arithmetic progression. Formally, let periodicity $\tau(A)$ be defined as: $$\tau(A) = \min \{s: a_{i+1}-a_i = a_{i+1+s}-a_{i+s} \text{ for all valid $i$ }\} $$ In words, this means elements of $A$ are periodic with a minimum cycle $\tau(A)$, up to some offset.
Now we choose one of two algorithms, according to whether $\tau(A)<n/3$ or not.
Case I ($\tau(A)<n/3$): Recursively compute $I=\mathcal{A}(A[1:s],B)$. Construct the new feasible index set $I'$ by selecting all indices $i\in I$ for which $i+s,i+2s,\dots, i+(n/s)\in I$. This holds because of the periodic pattern in $A$. This computation can be done by DP: Let $$R[i] = \lvert\{j\colon i+js\in I\}\rvert$$ Then, $R[i]$ can be computed based on $R[i+s]$ and $I$. $I'$ will correspond to indices for which $R[i]\ge n/s$. The computation of $I'$ will cost $O(m)$ overall.
Case II ($\tau(A)>n/3$): By definition, for $s=n/3$ there should be an index $i$ that $a_{i+1}-a_i \neq a_{i+1+s}-a_{i+s}$. If $i\le n/3$, we will have $i,i+s\le 2n/3$ which certifies that $\tau(A[1:2n/3])>n/3$. Otherwise, $i>n/3$ implies that $\tau(A[n/3:n])>n/3$.
Wlog assume $\tau(A[1:2n/3)>n/3$, and choose the lower half $A'=A[1:n/2]$ to recurse on (otherwise choose the upper half, same arguments would follow). Recursively compute $I=\mathcal{A}(A',B)$. For each index $i\in I$, check if the rest of the sequence can be found in $B$. Since both sequences are sorted this can be done in $O(n)$ step for each index, which implies an overall $O(|I|\cdot n)$ time to compute the valid indices, and return them as $\mathcal{A}(A, B)$. The efficiency of this step relies on the following claim:
Claim: $|I|\le 6m/n$, meaning that solutions are not too much overlapping.
Proof of claim: We show $|I|> 6m/n$ leads to a contradiction. Each index $i\in I$ is the starting point of a set of indices $J_i=\{i=j_1,\dots,j_{n/2}\}\subseteq B$, that map $A'$ to $B$ up to some offset. Collectively, there are at least $3m$ indices: $$\sum_{i\in I} |J_i| = |I|n/2\ge 6m/n \cdot n/2=3m$$ Since $|B|=m$, by pigeonhole principle, there is at least one index $x\in B$ appears in 3 separate solutions: $$\exists x\in B, r,s,p\in I\colon\; x\in J_r\cap J_s\cap J_p$$
Let $s $ be the median of the three $r<s<p$. Since $x\in J_s$, and $|J_s|=n/2$, $x$ partitions $J_s$ to two parts, one of which should have less than $n/4$ indices, which we assume is the lower part: $$J_s=\{j_1=s, j_2, \dots, j_\ell=x\}, \ell\le n/4$$ By construction, $s=j_1$ is mapped to $a_1$, up to $a_\ell$ up to some offset. But we also have $x\in J_p$, this implies a periodict less than $\ell\le n/4$ in $A'$, which contradicts $\tau(A')>n/3$. (there's a few details I will add later)
Overall Complexity At each step of the recursion, we pay $O(m)$. The periodicity $\tau(A)$ can also be computed in $O(n)$, by computing the longest suffix which is also prefix, of $diff(A)$, that is the increment array $A[2:n]-A[1:n-1]$. However, the size of the problem reduces by at least $1/2$ in each recursive step. There will be $\log n$ steps in worst case, which implies the time complexity is bounded by $O(m \log n)$. Adding the sorting costs, and since $m>n$, the overall complexity is bounded by the sorting time $O(m\log m)$
