Suppose Y is obtained by attaching a zero-cell, so $Y = X cup {ast}$. Then $Y^2$ is $$(X times X) cup (X times {ast}) cup ({ast} times X) cup ({ast} times {ast})$$
and so $Y^2/X^2$ is homeomorphic to
$$
{ast} cup X cup X cup {ast}.
$$
This can be arbitrarily complicated depending on X.
ADDENDUM: By request, a connected example is the inclusion $mathbb{CP}^1 subset mathbb{CP}^2$. The quotient $Y^2/X^2$, in this case, has a nonzero cohomology operation $Sq^2$ from H6 to H8 with ℤ/2-coefficients, and there is no wedge of products of spheres that can have this cohomology. (You should work out the details.)
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