To warm up, let's start with $k=3$. We want to prove that the numbers within the large square that's $nth$ from the left ad $mth$ from the top sum to
$$(2m-1)(2n-1)T_3^2= (2m-1)(2n-1)9.$$
First look at the large square that's $nth$ from the left and lies in the first row of large squares. As we already noted above, the two numbers in the top row of this white square are $3(n-1)+1=3n-2$ and $3(n-1)+2=3n-1.$
By some very similar reasoning we can work out that the top two numbers in the large square that's $nth$ from the left and $mth$ from the top are
$$\left(3(m-1)+1\right)\left(3n-2\right)=(3m-2)(3n-2)$$ and
$$\left(3(m-1)+1\right)\left(3n-1\right)=(3m-2)(3n-1)$$ Similarly, the bottom two numbers in the $nth$ white square from the left in the $mth$ row of white squares are
$$\left(3(m-1)+2\right)\left(3n-2\right)=(3m-1)(3n-2)$$ and
$$\left(3(m-1)+2\right)\left(3n-1\right)=(3m-1)(3n-1).$$
Adding these four numbers gives
$$\begin{array}{cc} & (3m-2)(3n-2) \\
+ &(3m-2)(3n-1) \\
+ & (3m-1)(3n-2) \\
+ & (3m-1)(3n-1) \end{array}$$
which we can rewrite as the product
$$\left(3m-2+3m-1\right)\left(3n-2+3n-1\right) = (6m-3)(6n-3)=(2m-1)(2n-1)9,$$
which is exactly what we wanted.
We now move on to prove the result for any positive integer $k$. We want to show that the sum of numbers in the square that's $nth$ from the left and $mth$ from the top is equal to
$$(2n-1)(2m-1)T_{k-1}^2.$$
Let's call the square in question $T.$
Using similar reasoning as above, you can see that the numbers in the first row of $T$ are
$$\left(k(m-1)+1\right)$$ times the numbers in the first row of $S$. This means that the sum of the first row of numbers in $T$ is
$$\left(k(m-1)+1\right)\left(2(n-1)T_{k-1}\right).$$
Similarly, the sum of the second row of numbers in $T$ is
$$\left(k(m-1)+2\right)\left(2(n-1)T_{k-1}\right).$$ We can continue in this vein until we come to the sum of the last row of numbers in $T,$ which is
$$\left(k(m-1)+(k-1)\right)\left(2(n-1)T_{k-1}\right).$$ Adding up these sums of the rows of $T$ gives
\begin{equation}\left[\left(k(m-1)+1\right)+ \left(\left(k(m-1)+2\right)+...+ \left(\left(k(m-1)+(k-1)\right)\right]\left[(2(n-1)T_{k-1}\right].\end{equation}
Similarly to what we did above, we can re-write this expression as
$$\left[1+2+...+(k-1) +(k-1)k(m-1)\right]\left[(2n-1)T_{k-1}\right].$$
Using the formula for the sum of the first $k-$ integers, this becomes
$$\left[\frac{(k-1)k}{2}+(k-1)k(m-1)\right](2n-1)T_{k-1},$$ which is equal to
$$\frac{(k-1)k}{2}(2m-1)(2n-1)T_{k-1}.$$ And since
$$\frac{(k-1)k}{2}=T_{k-1},$$
the sum of the numbers in $T$ is equal to
$$(2m-1)(2n-1)T_{k-1}^2.$$
This is what we wanted to show.