= Solution
Choose <orientations> on the two <sphere> factors whose product is the specified <orientation> of $X$. Give the four factors of $Y=(S^2)^4$ the corresponding <orientations> in the order $1,2,3,4$. The <Künneth theorem> has no <Tor functor> terms here and gives
$$
\boxed{H_j(Y;\mathbb Z)=
\begin{cases}
\mathbb Z,&j=0,8,\\
\mathbb Z^4,&j=2,6,\\
\mathbb Z^6,&j=4,\\
0,&\text{otherwise}.
\end{cases}}
$$
Let $u_i\in H^2(Y;\mathbb Z)$ be the pullback of the positively normalized <cohomology> class from factor $i$. The <cohomology ring> is
$$
H^*(Y;\mathbb Z)=\mathbb Z[u_1,u_2,u_3,u_4]/(u_1^2,u_2^2,u_3^2,u_4^2),
\qquad |u_i|=2,
$$
with $\langle u_1u_2u_3u_4,[Y]\rangle=1$. Its generators commute by <graded commutativity of the cup product>.
Fix a point in each <sphere>. For each pair $i<j$, let $S_{ij}$ be the four-dimensional <submanifold> in which factors $i,j$ vary and the remaining factors equal their fixed points. Orient $S_{ij}$ by the ordered factors $i,j$. The six <homology classes>
$$
[S_{12}],\ [S_{13}],\ [S_{14}],\ [S_{23}],\ [S_{24}],\ [S_{34}]
$$
give a <coordinate sphere intersection basis> by the <homology cross product> description in the <Künneth theorem>. Moving a fixed point in one of the other two factors produces a disjoint homologous representative of each $S_{ij}$. Hence every basis element has zero <self-intersection number>.
More precisely, the <intersection pairing> is
$$
[S_{ij}]\mathbin{\cdot}[S_{kl}]=
\begin{cases}
1,&\{i,j\}\sqcup\{k,l\}=\{1,2,3,4\},\\
0,&\text{otherwise}.
\end{cases}
$$
For complementary pairs the representatives intersect transversely at one point. The sign is positive because swapping two-dimensional factors contributes $(-1)^{2\cdot2}=1$. For pairs with a common index, a factor fixed in both representatives can be displaced to make them disjoint. In the paired order $(12,34),(13,24),(14,23)$, the <intersection form> is three copies of $\begin{pmatrix}0&1\\1&0\end{pmatrix}$.
For the <diagonal map> $\delta:X\to Y$, write $a,b\in H^2(X;\mathbb Z)$ for the two <sphere> generators. Then
$$
\delta^*u_1=\delta^*u_3=a,\qquad
\delta^*u_2=\delta^*u_4=b,
\qquad a^2=b^2=0,\quad\langle ab,[X]\rangle=1.
$$
The coefficient of $[S_{ij}]$ in $[\Delta]$ is $\langle u_iu_j,[\Delta]\rangle$. Pullback and the preceding <cup product> relations show that these coefficients are one for $12,14,23,34$ and zero for $13,24$. Therefore
$$
\boxed{[\Delta]=[S_{12}]+[S_{14}]+[S_{23}]+[S_{34}].}
$$
There are exactly two complementary pairings among these four terms, each counted twice in the square. Thus
$$
\boxed{[\Delta]\mathbin{\cdot}[\Delta]=2(1+1)=4.}
$$
This computes both the explicit <homology class> and its <intersection pairing>, including all <orientation> signs.
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