I think $[X] \frown \mathfrak{P}(m \oplus n) = 2\langle m \smile n, [Y] \rangle$.
The class $m \oplus n$ is better thought of as $m \otimes 1 + n \otimes x$ under the Kunneth decomposition, where $x \in H^1(S^1;\mathbb{Z}/2)$ is the nontrivial element. Then the quadratic property of $\mathfrak{P}$ and naturality gives $$\mathfrak{P}(m \otimes 1 + n \otimes x) = \mathfrak{P}(m) \otimes 1 + \mathfrak{P}(n \otimes x) + 2(m \smile n \otimes x).$$
Firstly, $\mathfrak{P}(m) = 0$ as $Y$ is 3-dimensional.
Secondly, consulting
Nakaoka, Minoru
Note on cohomological operations. J. Inst. Polytech.
Osaka City Univ. Ser. A. Math. 4, (1953). 51–58.
we find that $$\mathfrak{P}(n \otimes x) = \mathfrak{P}(n) \otimes \mathfrak{P}(x) + \bar{\mathfrak{P}}(n) \otimes \beta(Sq_2(x)) + \beta(Sq_2(n)) \otimes \bar{\mathfrak{P}}(n)$$ where $\bar{\mathfrak{P}}(-)$ is the operation given on cochains by $u \mapsto u \cup \delta u$, and $\beta$ is the Bockstein to $\mathbb{Z}/4$-cohomology. Each of $\mathfrak{P}(x)$, $\beta(Sq_2(x))$, and $\bar{\mathfrak{P}}(x)$ must be trivial by naturality (and degree reasons: $x$ is pulled back from a class on $S^1$), so $\mathfrak{P}(n \otimes x)=0$.