As others have said, this is a topological statement, not a smooth one.

This works for any two real vector spaces $V$ and $W$. They don't have to be two-dimensional.

  Let $A$ and $B$ be the unit spheres of $V$ and $W$. Let $\Delta$ be the closure, in the projective plane, of the quadrant $Q=[log\ 2,\infty)\times [log\ 2,\infty)\subset\mathbb R^2$. It is topologically a $2$-simplex. Map the product $A\times B\times\Delta$ to the one-point compactification of $V\times W$ by giving a proper map from the dense open subset $A\times B\times Q$ to $V\times W$, namely $$ (a,b,(s,t))\mapsto ((e^s-2)a,(e^t-2)b). $$ Map $A\times B\times\Delta$ also to the projective segment $\mathbb P^F$ by $$ (a,b,(s,t))\mapsto t/s. $$ The combined map $A\times B\times\Delta\to V\times W\times \mathbb P^F$ displays the space you are asking about as a quotient of $A\times B\times\Delta$. Now look at which points have been identified.

If $p\in \Delta$ belongs to the side of $\Delta$ at infinity, then $(a,b,p)$ is always identified with $(a',b',p)$. If $p$ belongs to the $t=log\ 2$ side then identification occurs when $b=b'$. If $p$ belongs to the $s=log\ 2$ side then identification occurs when $a=a'$.

If we just perform the last two identifications, we get as a quotient the triple join of $A$, $B$, and a point -- that is, the cone on the join $A\ast B$. The remaining identification makes the base of the cone into a single point. Thus the result is the (unreduced) suspension of the sphere $A\ast B$. This is again a sphere (of the same dimension as $V\times W$).

Not true. Let $S^1$ be the unit circle in $\mathbb C$. Let $\Delta$ be the closure, in the projective plane, of the quadrant $Q=[log\ 2,\infty)\times [log\ 2,\infty)\subset\mathbb R^2$. It is topologically a $2$-simplex. 

Map the product $S^1\times S^1\times\Delta$ to the one-point compactification of $\mathbb C\times \mathbb C$ by giving a proper map from the dense open subset $S^1\times S^1\times Q$ to $\mathbb C\times \mathbb C$, namely $$ (a,b,(s,t))\mapsto ((e^s-2)a,(e^t-2)b). $$ Map $S^1\times S^1\times\Delta$ also to the projective segment $\mathbb P^F$ by $$ (a,b,(s,t))\mapsto t/s. $$ The combined map $S^1\times S^1\times\Delta\to \mathbb C\times \mathbb C\times \mathbb P^F$ displays the space you are asking about as a quotient of $S^1\times S^1\times\Delta$. Now look at which points have been identified. For a point $p$ in the interior of the ``infinity'' side of $\Delta$, $S^1\times S^1\times p$ goes to one point, and a neighborhood of that point in the quotient space looks like the product of an open interval and a cone on $S^1\times S^1$. Thus a neighborhood looks like a cone on the suspension of $S^1\times S^1$, and is not a manifold.