@hexomino has found optimal solutions with a restricted search space.
Theorem:
There is a solution with as large a sum of sack-labels as possible that doesn't put more than 2 coins into any bag.
Proof:
Suppose the number of sacks containing exactly $k$ coins is $c_k$. Then it's easy to see that the sum of sack-labels is $\sum_{i<j}c_ic_j=\frac12\sum_{i\neq j}c_ic_j=\frac12\left(\left(\sum c_k\right)^2-\sum c_k^2\right)$. Now, suppose we have at least one sack with $k>2$ coins in it. What happens if we take one of those coins out and put it in its own sack? Well, write $n:=\sum c_k$ for the total number of sacks. This increases by 1. At the same time $c_k$ decreases by 1 and $c_1,c_{k-1}$ increase by 1. Note that $1$ and $k-1$ aren't equal since $k>2$. So, $\left(\sum c_k\right)^2=n^2$ increases by $2n+1$; $c_k^2$ decreases by $2c_k-1$; $c_{k-1}^2$ increases by $2c_{k-1}+1$; and $c_1^2$ increases by $2c_1+1$. Hence $\frac12\left(\left(\sum c_k\right)^2-\sum c_k^2\right)$ increases by $\frac12\left((2n+1)+(2c_k-1)-(2c_{k-1}+1)-(2c_1+1)\right)=n+c_k-c_{k-1}-c_1-1$. And $n\geq c_{k-1}+c_1$ because $n$ is the sum of the $c$'s; and $c_k\geq1$ because we assumed we do have a sack of size $k$ to take a coin out of. Hence, our sack-label-sum doesn't decrease when we make this change. So we can keep doing it until we no longer have any sacks with $>2$ coins in, and the sack-label-sum will be at least as big as the one we started with. (Once we are in that situation, the argument above doesn't say that we should keep splitting sacks, because it's invalidated by the fact that when $k=2$ we have $k-1=1$.)
Therefore
@hexomino's solution that assumes no sack has more than 2 coins in it is in fact optimal. (For the sum-of-labels case; I haven't looked at the product-of-labels case.)
Theorem
There is a solution with as large a product of sack-labels as possible that doesn't put more than 2 coins in any sack.
Proof
Same strategy as before: consider turning a sack with $k$ coins into a sack with $k-1$ coins and a new sack with $1$ coin. Let's look in more details at the labels, and for simplicity let's suppose that the sack we do this to is a largest sack: there are no others containing more coins. Sacks with $1$ coin are unlabelled and don't change. Sacks whose coin-count is $>1$ have a label $1$ larger than before because of the new size-$1$ sack EXCEPT that (1) any unaltered sacks of size $k$ have also gained a new size-$k-1$ sack to be bigger than, so their labels increase by $2$ instead of by $1$ and, alas (2) our new sack of size $k-1$ may suffer a decrease in its label, if there were sacks of size $k-1$ that it's no longer bigger than. So, suppose we had $p$ sacks of size $1$, $q$ sacks of size $2\dots k-2$, $r$ sacks of size $k-1$, and $s$ sacks of size $k$. Then $p$ unlabelled sacks become $p+1$ unlabelled sacks; $q$ "small" sacks have their labels increased by $1$; $r$ $k-1$-sacks also have their labels increased by 1; one reduced-size sack has its label reduced by $r-1$; the other $s-1$ "large" sacks have their labels increased by $2$. And, of course, larger sacks have larger labels. In particular, the product of (old) size-$k-1$ sacks' labels has changed by $\left(\frac{l+1}{l}\right)^r$ where $l$ is the label that used to be on those sacks, and the only label-decrease is by a factor of $\frac{l'-r+1}{l'}$ where $l'$ is the label that used to be on sacks of size $k$. Well, in fact we know that $l'=l+r$. So the overall change in label-product is $\left(\frac{l+1}l\right)^r\frac{l+1}{l+r}$ times the changes arising from sacks that don't end up having size $k-1$, and those latter are always increases. Writing $h:=1/l$, that fraction equals $\frac{(1+h)^{r+1}}{1+rh}$. This is $1$ when $h=0$ and its derivative w.r.t. $h$ turns out to be $\frac{(1+h)^r(1+r^2h)}{(1+rh)^2}$ which is positive. So the fraction's value is $\geq1$ for any $h>0$, and in particular is $\geq1$ when $h=1/l$, and therefore our change doesn't decrease the product of the labels. And now we are done, as before.
Therefore
@hexomino's solution for products is optimal across the whole space of possible sack-sizes.
