I think that you are correct in the hint that you're remembering that the exercise told you to consider.
My idea is that, using this hint, you ought to be able to proceed inductively in constructing the regular complex $Y$.
Start by looking at all attaching maps $\varphi_{\alpha}:S^{\alpha} \rightarrow X_0$ onto the $0$-skeleton for $X$. Since $X$ is finite, these can be listed: $\{\varphi_{\alpha_1},...,\varphi_{\alpha_r}\}$.
Now, as you suggest, for each of these we wish to somehow consider instead the map $S^{\alpha} \rightarrow X_0 \times D^{\alpha+1}$, because this would be injective. What we therefore do is, for each $\alpha_1,...,\alpha_r$, take the successive product of $X_0$ with each of the $D^{{\alpha_i}+1}$s. This results in a space
$$Y_0 := X_0 \times D^{{\alpha_1}+1} \times....\times D^{{\alpha_r}+1}.$$
We can now use the standard cellular decomposition of $D^{k+1}$ to obtain a cellular decomposition of the product above. This is the "bottom" of our new regular complex $Y$. It is not the $0$-skeleton for our CW-complex $Y$, but it is homotopy equivalent to the $0$-skeleton of $X$. As such, we do indeed mildly abuse notation by calling this thing $Y_0$.
We then proceed with attaching cells to this product via the maps $\psi_{\alpha}$ as you suggest above. Importantly, we proceed as we normally would when we construct a CW-complex. We first attach the $1$-cells, and procede with $2$-cells, etc.
It is important to note a couple of things at this point.
We have certainly dealt with all attaching maps
for the $1$-skeleton of $X$ at this point. What this means is that
this process can indeed now be iterated; we look at attaching maps onto $X_1$, take whatever products we need to take with disks D^{\alpha}$ in order that these maps will be injective, and continue.
Importantly what we have constructed (and what we construct inductively in
subsequent steps) is homotopy equivalent to $X_0$, $X_1$ and so on.
It is indeed easy enough to see that the product $X_0 \times
D^{{\alpha_1}+1} \times....\times D^{{\alpha_r}+1}$ is homotopy
equivalent to $X_0$, but what about the next steps? We wish to see that $Y_1$ is homotopy equivalent to $X_1$. This can be seen in a few different ways, but the simplest for these purposes is just to state the following lemma (which I state rather wordily):
Lemma: Due to the nature of the construction of CW-complexes,
whatever space $X$ you're attaching a particular cell to can be replaced by a
homotopy equivalent space $\overline{X}$ (and the attaching map by the appropriate homotopic map) and the space which results $\overline{X} \cup D^{k}$ is homotopy equivalent to $X \cup D^k$.
Now what we're essentially doing is replacing $X_0$ with the homotopy equivalent $Y_0$, and attaching each $1$-cell to this new thing. Utilising the above lemma for each cell we attach gives that the resulting space is homotopy equivalent to $X_1$. We now "fatten up" this resulting space in exactly the same manner as how we "fattened up" $X_0$, by taking products with disks. Taking products with disks trivially results in a homotopy equivalent space, and what we end up with is the space which we will label $Y_1$; homotopy equivalent to $X_1$.
I think that iteratively doing this should complete the proof.
I should note that whilst the $Y_0$ and $Y_1$ which I mention above are certainly not the $0$ or $1$ skeletons for the space $Y$, there is nevertheless a natural $CW$-complex structure available for $Y$ which is obtained by using the standard cellular decomposition of the disks $D^k$, and this is then indeed a regular complex.
Separate $S^1$ in four arcs call them $p_1p_2,p_2p_3,p_3p_4,p_4p_1$ and direction is clockwise and identify $D^2$ as $I^2$ , mark this square square as ABCD in an anticlockwise direction starting from left.
Let us also call the two circles of $S^1\vee S^1$ by $a,b$ with appropriate direction (think them as a directed path).
Now you map the arc $\overrightarrow{p_1p_2}$
to side $\overrightarrow{AB}$ of the square , the arc $\overrightarrow{p_2p_3}$
to side $\overrightarrow{BC}$ of the square and so on.
Now, we give the identification map as follows, map the arc $\overrightarrow{p_1p_2}$ to the directed loop $a$ then the arc $\overrightarrow{p_2p_3}$ to the directed loop $b$ , then the arc $\overrightarrow{p_3p_4}$ to the directed loop $a^{-1}$ (i.e. now traverse in the opposite direction) and finally the arc $\overrightarrow{p_4p_1}$ to the directed loop $b^{-1}$ and this gives you the torus as the initial object of the pushout diagram
Best Answer
I don't know why someone would say that the attaching map of $e_m \times e_n$ has the form $\phi_\alpha \times \psi_\beta$. That would seem to imply that the domain of the attaching map is $S^{m-1} \times S^{n-1}$, which does not even have the right dimension for the boundary of a cell of dimension $m+n$.
Instead the boundary of $D^m \times D^n$ is $(S^{m-1} \times D^n) \cup (D^m \times S^{n-1})$. You can convince yourself that this is homeomorphic to $S^{m+n-1}$, via the restriction of some homeomorphism from $D^m \times D^n$ to $D^{m+n}$.
So the attaching map for $e^m_\alpha \times e^n_\beta$ must be a function of the form $$\gamma_{\alpha,\beta} : (S^{m-1} \times D^n) \cup (D^m \times S^{n-1}) \to (X \times Y)^{(m+n-1)} $$ We already have attaching maps for the cells $e^m_\alpha$ and $e^n_\beta$ of the form $$\phi_\alpha : S^{m-1} \to X^{(m-1)} \qquad\qquad \psi_\beta : S^{n-1} \to Y^{(n-1)} $$ which extend to characteristic maps for those cells of the form $$\chi_\alpha : D^m \to X^{(m)} \qquad\qquad \omega_\beta : D^n \to Y^{(n)} $$ The definition of the attaching map for $e^m_\alpha \times e^n_\beta$ can therefore be given by the function $$\gamma_{\alpha,\beta}(x,y) = \begin{cases} (\phi_\alpha(x),\omega_\beta(y)) & \quad\text{if $(x,y) \in S^{m-1} \times D^n$} \\ (\chi_\alpha(x),\psi_\beta(y)) &\quad\text{if $(x,y) \in D^m \times S^{n-1}$} \end{cases} $$ and one should note that $$(\phi_\alpha(x),\omega_\beta(y)) \in X^{(m-1)} \times Y^n \subset (X \times Y)^{m+n-1} $$ and that $$(\chi_\alpha(x),\psi_\beta(y)) \in X^m \times Y^{n-1} \subset (X \times Y)^{m+n-1} $$
Seeing the formula for $\gamma_{\alpha,\beta}$, one could say that $\gamma_{\alpha,\beta}$ is the restriction to $(S^{m-1} \times D^n) \cup (D^m \times S^{n-1})$ of the product of the characteristic maps $\chi_\alpha \times \omega_\beta$.
But it is certainly wrong to say it is the product of the attaching maps $\phi_\alpha \times \psi_\beta$. I'm curious to know where you saw such answers.