There's no problem with summing over an uncountable indexing set, but over the reals you don't really get anything interesting by moving from the countable case to the uncountable case.

Suppose we have such an uncountable summation of nonnegative real numbers: $\sum_{r \in \Gamma} x_r$. Then we have a few cases:

(i) One of the $x_r = \infty$. Then our sum, $\sum_{r \in \Gamma} x_r$, is also $\infty$.

(ii) Suppose none of the $x_r =\infty$. Now we utilize dyadic decomposition to write the positive reals as a countable union of sets: $(0, \infty] = \cup_{j \in \mathbb{Z}} (2^j, 2^{j+1}]$.
By the pigeonhole principle, either (a) There is a $j$ such that there are uncountably many nonzero $x_r$ in $(2^j, 2^{j+1}]$, in which case our sum is $\infty$ again or (b) there are only countably many nonzero $x_r$ in each interval for all $j$. In the latter case, we're back to our countable sum, as a countable union of countable sets is still countable.

Now, if our indexing set $\Gamma$ is essentially countable in this way, we can motivate our sum as usual by defining $\sum_{r \in \Gamma} x_r = \sup_{E \subset \Gamma} \sum_{r \in E} x_r$, where $E$ is a finite set. So an uncountable sum can be defined in the same way, without really picking up many problems (or interesting cases, sadly).