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There was a gap in my first answer to the first question, discussed in the comments.

First question: I interpret 'symmetry' to mean 'quotienting by a finite group action on $S^3$, not necessarily free'; perhaps more strongly I interpret this as meaning 'quotient by some involution' (since you mention two 3-cells). Then the Euler characteristic of the result of attaching a 4-cell to $S^3/\iota$ is $\chi(S^3/\iota) + 1$. Write the fixed set as $F \subset S^3$; then $\chi(S^3/\iota) = \chi(F) + 1/2 \chi(S^3, F) = \chi(S^3) + 1/2 \chi(F)$. Thus $\chi(M) = 1 + 1/2 \chi(F)$. Now by Smith theory the fixed set of any involution on the 3-sphere has cohomology of rank at most 2, so that $\chi(M) \le 2$. We cannot obtain $\Bbb{CP}^2$ in this way.

I think the stronger claim is still not true, but I haven't attempted to prove it.

But if you allow the quotient to be done in a non-cellular manner, you're in luck. There's a uniform statement for $k = \Bbb R, \Bbb C, \Bbb H$ (and a slightly more careful discussion for $k = \Bbb O$). Take the unit disc $D(k^2)$ in $k^2$; its boundary is a sphere, which is acted on by the groups $S(k)$ of unit-norm elements in the respective algebras ($S^0, S^1, S^3$ respectively). Then $k\Bbb P^2$ is homeomorphic to $D(k^2)$ modulo the action of $S(k)$ on $\partial D(k^2) = S(k^2)$. (Exercise left to the reader.)

For $k = \Bbb R$, this recovers your description: $\Bbb{RP}^2$ is identified with $D^2$ modulo the antipodal action on the boundary, which is what gives you the "cell-pasting" description you're after.

There was a gap in my first answer to the first question, discussed in the comments. I hope it's clear that this doesn't constitute a full answer to OP's question, which remains interesting.

First question: I interpret 'symmetry' to mean 'quotienting by a finite group action on $S^3$, not necessarily free'; perhaps more strongly I interpret this as meaning 'quotient by some involution' (since you mention two 3-cells). Then the Euler characteristic of the result of attaching a 4-cell to $S^3/\iota$ is $\chi(S^3/\iota) + 1$. Write the fixed set as $F \subset S^3$; then $\chi(S^3/\iota) = \chi(F) + 1/2 \chi(S^3, F) = \chi(S^3) + 1/2 \chi(F)$. Thus $\chi(M) = 1 + 1/2 \chi(F)$. Now by Smith theory the fixed set of any involution on the 3-sphere has cohomology of rank at most 2, so that $\chi(M) \le 2$. We cannot obtain $\Bbb{CP}^2$ in this way.

I think the stronger claim is still not true, but I haven't attempted to prove it.

But if you allow the quotient to be done in a non-cellular manner, you're in luck. There's a uniform statement for $k = \Bbb R, \Bbb C, \Bbb H$ (and a slightly more careful discussion for $k = \Bbb O$). Take the unit disc $D(k^2)$ in $k^2$; its boundary is a sphere, which is acted on by the groups $S(k)$ of unit-norm elements in the respective algebras ($S^0, S^1, S^3$ respectively). Then $k\Bbb P^2$ is homeomorphic to $D(k^2)$ modulo the action of $S(k)$ on $\partial D(k^2) = S(k^2)$. (Exercise left to the reader.)

For $k = \Bbb R$, this recovers your description: $\Bbb{RP}^2$ is identified with $D^2$ modulo the antipodal action on the boundary, which is what gives you the "cell-pasting" description you're after.

There was a gap in my first answer to the first question, discussed in the comments.

First question: I interpret 'symmetry' to mean 'quotienting by a finite group action on $S^3$, not necessarily free'; perhaps more strongly I interpret this as meaning 'quotient by some involution' (since you mention two 3-cells). Then the Euler characteristic of the result of attaching a 4-cell to $S^3/\iota$ is $\chi(S^3/\iota) + 1$. Write the fixed set as $F \subset S^3$; then $\chi(S^3/\iota) = \chi(F) + 1/2 \chi(S^3, F) = \chi(S^3) + 1/2 \chi(F)$. Thus $\chi(M) = 1 + 1/2 \chi(F)$. Now by Smith theory the fixed set of any involution on the 3-sphere has cohomology of rank at most 2, so that $\chi(M) \le 2$. We cannot obtain $\Bbb{CP}^2$ in this way. (Notice that $\chi(M) = 2$ remains possible, as $M = \Bbb{RP}^4$ does arise in this way!)

I think the stronger claim is still not true, but I haven't attempted to prove it.

But if you allow the quotient to be done in a non-cellular manner, you're in luck. There's a uniform statement for $k = \Bbb R, \Bbb C, \Bbb H$ (and a slightly more careful discussion for $k = \Bbb O$). Take the unit disc $D(k^2)$ in $k^2$; its boundary is a sphere, which is acted on by the groups $S(k)$ of unit-norm elements in the respective algebras ($S^0, S^1, S^3$ respectively). Then $k\Bbb P^2$ is homeomorphic to $D(k^2)$ modulo the action of $S(k)$ on $\partial D(k^2) = S(k^2)$. (Exercise left to the reader.)

For $k = \Bbb R$, this recovers your description: $\Bbb{RP}^2$ is identified with $D^2$ modulo the antipodal action on the boundary, which is what gives you the "cell-pasting" description you're after.

There was a gap in my first answer to the first question, discussed in the comments.

First question: I interpret 'symmetry' to mean 'quotienting by a finite group action on $S^3$, not necessarily free'; perhaps more strongly I interpret this as meaning 'quotient by some involution' (since you mention two 3-cells). Then the Euler characteristic of the result of attaching a 4-cell to $S^3/\iota$ is $\chi(S^3/\iota) + 1$. Write the fixed set as $F \subset S^3$; then $\chi(S^3/\iota) = \chi(F) + 1/2 \chi(S^3, F) = \chi(S^3) + 1/2 \chi(F)$. Thus $\chi(M) = 1 + 1/2 \chi(F)$. Now by Smith theory the fixed set of any involution on the 3-sphere has cohomology of rank at most 2, so that $\chi(M) \le 2$. We cannot obtain $\Bbb{CP}^2$ in this way.

I think the stronger claim is still not true, but I haven't attempted to prove it.

But if you allow the quotient to be done in a non-cellular manner, you're in luck. There's a uniform statement for $k = \Bbb R, \Bbb C, \Bbb H$ (and a slightly more careful discussion for $k = \Bbb O$). Take the unit disc $D(k^2)$ in $k^2$; its boundary is a sphere, which is acted on by the groups $S(k)$ of unit-norm elements in the respective algebras ($S^0, S^1, S^3$ respectively). Then $k\Bbb P^2$ is homeomorphic to $D(k^2)$ modulo the action of $S(k)$ on $\partial D(k^2) = S(k^2)$. (Exercise left to the reader.)

For $k = \Bbb R$, this recovers your description: $\Bbb{RP}^2$ is identified with $D^2$ modulo the antipodal action on the boundary, which is what gives you the "cell-pasting" description you're after.

First question: no. If the 4-ball is divided into two 3-cells, they intersect along their boundary (a 2-sphere), and you've produced a CW-complex with one 0-cell, one 2-cell, one 3-cell, and one 4-cell. This has the wrong Euler characteristic to be $\Bbb{CP}^2$. I'm skeptical about your second phrasing, too, but didn't attempt a proof that it's impossible.

But if you allow the quotient to be done in a non-cellular manner, you're in luck. There's a uniform statement for $k = \Bbb R, \Bbb C, \Bbb H$ (and a slightly more careful discussion for $k = \Bbb O$). Take the unit disc $D(k^2)$ in $k^2$; its boundary is a sphere, which is acted on by the groups $S(k)$ of unit-norm elements in the respective algebras ($S^0, S^1, S^3$ respectively). Then $k\Bbb P^2$ is homeomorphic to $D(k^2)$ modulo the action of $S(k)$ on $\partial D(k^2) = S(k^2)$. (Exercise left to the reader.)

For $k = \Bbb R$, this recovers your description: $\Bbb{RP}^2$ is identified with $D^2$ modulo the antipodal action on the boundary, which is what gives you the "cell-pasting" description you're after.

There was a gap in my first answer to the first question, discussed in the comments.

First question: I interpret 'symmetry' to mean 'quotienting by a finite group action on $S^3$, not necessarily free'; perhaps more strongly I interpret this as meaning 'quotient by some involution' (since you mention two 3-cells). Then the Euler characteristic of the result of attaching a 4-cell to $S^3/\iota$ is $\chi(S^3/\iota) + 1$. Write the fixed set as $F \subset S^3$; then $\chi(S^3/\iota) = \chi(F) + 1/2 \chi(S^3, F) = \chi(S^3) + 1/2 \chi(F)$. Thus $\chi(M) = 1 + 1/2 \chi(F)$. Now by Smith theory the fixed set of any involution on the 3-sphere has cohomology of rank at most 2, so that $\chi(M) \le 2$. We cannot obtain $\Bbb{CP}^2$ in this way. (Notice that $\chi(M) = 2$ remains possible, as $M = \Bbb{RP}^4$ does arise in this way!)

I think the stronger claim is still not true, but I haven't attempted to prove it.

But if you allow the quotient to be done in a non-cellular manner, you're in luck. There's a uniform statement for $k = \Bbb R, \Bbb C, \Bbb H$ (and a slightly more careful discussion for $k = \Bbb O$). Take the unit disc $D(k^2)$ in $k^2$; its boundary is a sphere, which is acted on by the groups $S(k)$ of unit-norm elements in the respective algebras ($S^0, S^1, S^3$ respectively). Then $k\Bbb P^2$ is homeomorphic to $D(k^2)$ modulo the action of $S(k)$ on $\partial D(k^2) = S(k^2)$. (Exercise left to the reader.)

For $k = \Bbb R$, this recovers your description: $\Bbb{RP}^2$ is identified with $D^2$ modulo the antipodal action on the boundary, which is what gives you the "cell-pasting" description you're after.