In this context, $\otimes$ refers specifically to the Kronecker product. In particular, we have
$$
I_d \otimes B
= \overbrace{B \oplus B \oplus \dots \oplus B}^d
= \text{diag}(\overbrace{B,B, \dots, B}^d)\\
= \pmatrix{B\\&B\\&&\ddots\\&&&B}
$$
Write $\|x\|^2=\sum_i c_i\langle x,u_i\rangle^2$ in matrix form as
$$
x'x=\sum_i c_i x'u_iu'_ix,
$$
where the prime indicates transposition. Set $x'=t'_j$, the $j^\text{th}$ row of $T$. So
$$
t_j't_j=\sum_i c_i t'_ju_iu'_it_j.
$$
Summing $t'_ju_iu'_it_j$ over $j$ gives $\|Tu_i\|^2$. Therefore
$$
\sum_j\|t_j\|^2=\sum_ic_i\|Tu_i\|^2.
$$
Now $\sum_j\|t_j\|^2\ge n$ when $\det T=1$, while $\sum_i c_i=n$ and all $c_i$ are positive. This gives a contradiction if all $\|Tu_i\|$ are less than $1$. Note that the sum over $\|t_j\|^2$ is the square of the Hilbert–Schmidt or Frobenius norm of $T$.
We should explain where the properties just invoked come from. That $\sum_j\|t_j\|^2\ge n$ for $\det T=1$ should be intuitively clear: $\det T$ is the volume of an $n$-dimensional parallelepiped. For given side lengths of the parallelepiped, the volume is maximized when all sides are orthogonal, while the sum of the squares of the side lengths of an $n$-dimensional rectangular prism of fixed volume is minimized when all sides are equal (as can be shown, for example, by the method of Lagrange multipliers). So for unit volume, the parallelepiped that minimizes the sum of the squares of the side lengths is the unit cube, for which $\sum_j\|t_j\|^2=n$.
That the $c_i$ are positive is one of the given conditions. The condition $\sum_i c_i=n$ can be obtained by letting $x$ in the second condition equal each of the standard basis vectors of $\mathbf{R}^n$ in turn, and summing:
$$
n=\sum_j e'_j e_j=\sum_j\sum_ic_ie'_ju_iu'_ie_j=\sum_ic_i\|u_i\|^2=\sum_ic_i.
$$
I'm not sure, unfortunately, that I can say much about your original question (about the meaning of "close to a proper subspace"). I do think it's worth working out a variety of detailed examples, even in dimension $2$. For a rhombus, which is what Figure 14 in Ball's article shows, the condition fails, unless the rhombus is a square. The same is true, I believe, for rectangles, and parallelograms more generally. You could, for example, let the four points in Figure 14 where the inscribed circle touches the rhombus be $\left(\pm\frac{5}{13},\pm\frac{12}{13}\right)$. Then pick some random points for $x$—the points $(1,0)$ and $(0,1)$ will do— and try to solve for the $c_i$. You will gain an appreciation for why there is no solution.
So if parallelograms don't work (except for squares), try hexagons. You can cut off the two corners in Figure 14 on the horizontal axis by adding vertical line segments tangent to the circle at $(-1,0)$ and $(1,0)$ to form a hexagon. Now solve the modified system of equations. What allows this solution in this case is the addition of the two vectors $(-1,0)$ and $(1,0)$, which, unlike the other four vectors, are not close to the subspace along the vertical axis. Now if the two vertical sides were moved outward by a little bit, so that the circle was no longer tangent to those sides, we would be back in a situation similar to the rhombus.
Best Answer
It's the Outer/Tensor product and defined by $u \otimes v = uv^\top$ or alternatively, you can think of it as the function
$$ (u \otimes v)(x) = \langle v, x \rangle u, $$
which is of course what you get when you multiply $uv^\top$ and $x$ together.
Note: there is also a Kronecker product which uses the same symbol and the relationship is
$$ u \otimes_{\rm out} v = u \otimes_{\rm kr} v^\top. $$
The Kronecker product of two column vectors $(a_i), (b_j)$ is a vector whose $(in + j)$-th entry is $a_ib_j$. The outer product is a matrix whose $(i,j)$-th entry is $a_ib_j$.