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 $\mathfrak{P}(n \otimes x)=0$.

I think $[X] \frown \mathfrak{P}(m \oplus n) = 2\langle m \cdot n + n^3, [Y] \rangle$.

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, the fact that the suspension of the Pontrjagin square is the Postnikov square (and that the Postnikov square is not universally trivial, which bizarrely I can't find a reference for) means that $$\mathfrak{P}(n \otimes x) = \bar{\mathfrak{P}}(n) = 2 n^3.$$

Remark: In an earlier version of this answer I had consulted

Nakaoka, Minoru
Note on cohomological operations. J. Inst. Polytech.
Osaka City Univ. Ser. A. Math. 4, (1953). 51–58.

which has the formula $$\mathfrak{P}(n \otimes x) = \mathfrak{P}(n) \otimes \mathfrak{P}(x) + \bar{\mathfrak{P}}(n) \otimes \beta(Sq_2(x)) + \beta(Sq_2(n)) \otimes \bar{\mathfrak{P}}(n)$$ where $\bar{\mathfrak{P}}(-)$ is the Postnikov square (i.e. the operation given on cochains by $u \mapsto u \cup \delta u$), and $\beta$ is the Bockstein to $\mathbb{Z}/4$-cohomology. Each of $\mathfrak{P}(x)$ and $\bar{\mathfrak{P}}(x)$ must be trivial by degree reasons. If one interprets $Sq_2(x)$ literally it also ought to be zero, but this is apparently wrong and it ought to be interpreted as $1$.

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, the class $\mathfrak{P}(n \otimes x)$ is zero because $n \otimes x$ is classified by a map to $K(\mathbb{Z}/2,1) \times K(\mathbb{Z},1)$ (as $x$ is the reduction of an integral class) and so $\mathfrak{P}(n \otimes x)$ is pulled back by the classifying map from the class $$\mathfrak{P}(\iota_2 \otimes \rho_2(\iota_1)) \in H^4(K(\mathbb{Z}/2,1) \times K(\mathbb{Z},1);\mathbb{Z}/4) = \mathbb{Z}/2$$ which is either zero or $\beta(\iota_2)^2 \otimes 1$ (I'm not sure). But in any case $n \otimes x$ pulls this back to zero or $\beta(n)^2 \otimes 1$, which is also zero as $Y$ is 3-dimensional.

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 $\mathfrak{P}(n \otimes x)=0$.