Is it possible to swap sums like that?

Not at all, look:

$$\sum_{a=1}^2\sum_{b=1}^3 \frac ab=\sum_{a=1}^2 \left(\frac a1+\frac a2+\frac a3\right)=\frac 11+\frac 12+\frac 13+\frac 21+\frac 22+\frac 23=\frac {11}2,$$

and

$$\sum_{a=1}^3\sum_{b=1}^2 \frac ab=\sum_{a=1}^3 \left(\frac a1+\frac a2\right)=\frac 11+\frac 12+\frac 21+\frac 22+\frac 31+\frac 32=9.$$

You can swap the sums, but not just what is on the top of your sum.

You do have for all finite sums:

$$\sum_{a=0}^n\sum_{b=0}^m f_{ab}=\sum_{b=0}^m\sum_{a=0}^n f_{ab}.$$

Edit

What you want to prove works because you are dealing with finite sums , and you changed the indices too.

Your illustration is fine:

$\sum_{z=0}^k\sum_{u=0}^j A_{\alpha u}B_{uz}C_{zb}=$

$\sum_{u=0}^k\sum_{z=0}^j A_{\alpha z}B_{zu}C_{ub}$

That's just swapping out the variables $u$, and $z$. The variables are labels so it doesn't matter what we call them.

Then $\sum_{u=0}^k\sum_{z=0}^j A_{\alpha z}B_{zu}C_{ub}=$

$\sum_{z=0}^j\sum_{u=0}^k A_{\alpha z}B_{zu}C_{ub}$

This is fine as addition is commutative so it doesn't matter in what order you add them. (Should be careful about infinite sums however. Infinite sums and grouping abuse can lead to "cheating" paradoxes".)

But $ \sum_{a=0}^n\sum_{b=0}^m f_{ab} \ne \sum_{b=0}^n\sum_{c=0}^m f_{ab}$

On the LHS you are adding $b$ up to $m$ (presumably a set number) and $a$ up to $n$. On the RHS you are adding $b$ up to $n$ and $a$ up to $m$. These are completely different results.

However the following are true.

$\sum_{a=0}^n\sum_{b=0}^m f_{ab} = \sum_{b=0}^m\sum_{a=0}^n f_{ab}=$

$ \sum_{b=0}^n\sum_{a=0}^m f_{ba} = \sum_{a=0}^m\sum_{b=0}^n f_{ba}$

No matter how you label $f_{ij}$ the first $i$ or $j$ goes to $n$ and the second $i$ or $j$ goest to $m$ and the order of which you add first doesn't matter.

We have $$\sum_{a=0}^1\sum_{b=0}^2 (a+2b)=\sum_{a=0}^1 (3a+6)=6+9=15,\quad \sum_{a=0}^2\sum_{b=0}^1 (a+2b)=\sum_{a=0}^2 (2a+2)=2+4+6=12,$$ so it is different.