It is a bit unclear from your post whether you are interested in the case where summands are non-negative real numbers or you want the more general case where summands belong to some topological vectors space.
Real numbers
In the case of non-negative real numbers we have $$\sum\limits_{\lambda\in\Lambda} c_\lambda = \sup \{\sum\limits_{\lambda\in\Lambda_0} c_\lambda; \Lambda_0\subseteq\Lambda; \Lambda_0\text{ is finite}\}.$$
In this case it is pretty straightforward to see that
$$\sum_{\lambda\in\Lambda} \sum_{\kappa \in K} a_{\lambda,\kappa} = \sup \{\sum\limits_{(\lambda,\kappa)\in\Lambda_0} a_{\lambda,\kappa}; \Lambda_0\subseteq\Lambda\times K; \Lambda_0\text{ is finite}\} = \sum_{\kappa \in K} \sum_{\lambda\in\Lambda} a_{\lambda,\kappa}$$
For the addition we have for each $\Lambda_0\subseteq\Lambda$
$$\sum_{\lambda\in\Lambda_0} (a_\lambda+b_\lambda) = \sum_{\lambda\in\Lambda_0} a_\lambda + \sum_{\lambda\in\Lambda_0} b_\lambda \le \sum_{\lambda\in\Lambda} a_\lambda + \sum_{\lambda\in\Lambda} b_\lambda $$
which implies
$$\sum_{\lambda\in\Lambda} (a_\lambda+b_\lambda) \le \sum_{\lambda\in\Lambda} a_\lambda + \sum_{\lambda\in\Lambda} b_\lambda.$$
To get the opposite direction we notice that for $\Lambda_0=\Lambda_1\cup\Lambda_2$ we get
$$\sum_{\lambda\in\Lambda_1} a_\lambda + \sum_{\lambda\in\Lambda_2} b_\lambda \le \sum_{\lambda\in\Lambda_0} (a_\lambda+b_\lambda) \le \sum_{\lambda\in\Lambda} (a_\lambda+b_\lambda)$$
which implies
$$\sum_{\lambda\in\Lambda} a_\lambda + \sum_{\lambda\in\Lambda} b_\lambda \le \sum_{\lambda\in\Lambda_0} (a_\lambda+b_\lambda) \le \sum_{\lambda\in\Lambda} (a_\lambda+b_\lambda).$$
Topological vector space
If you want to show additivity for more general sums in topological vector space $X$, you can use the fact that $+$ is a continuous map from $X\times X$ to $X$ and that continuous maps preserve convergence of nets.
This result can be also found as Theorem 9.1.2 in Dixmier's General topology.
The second result is not true in arbitrary topological vector spaces. Just consider $\Lambda=K=\mathbb N$ and $a_{n,n}=1$, $a_{n+1,n}=-1$ and other terms zero.
$$
\begin{array}{cccccc}
1 & 0 & 0 & 0 & \dots \\
-1 & 1 & 0 & 0 & \dots \\
0 &-1 & 1 & 0 & \dots \\
0 & 0 &-1 & 1 & \dots \\
\vdots & & & \ddots & \ddots
\end{array}
$$
All vertical sums are zeroes, but one of the horizontal sums is equal to one.
However, it is true at least in Banach spaces if we add the assumption that the sum $\sum\limits_{(\kappa,\lambda)\in K\times\Lambda} a_{\kappa,\lambda}$ exists.
I found in Dixmier's General Topology Theorem 9.2.2 which says this:
Theorem 9.2.2 (Associativity)
Let $E$ be a Banach space, $(x_i)_{i\in I}$ a summable family in $E$, $(I_l)_{l\in L}$ a partition of $I$. For every $l\in L$, set $y_i=\sum_{i\in I}$, which is meaningful by 9.2.1. Then the family $(y_l)_{l\in L}$ is summable and $\sum_{i\in I}x_i=\sum_{l\in L} y_l$.
The Theorem 9.2.1 mentioned there says that if some family in a Banach space is summable, then a subfamily is summable, too.
The completeness is used in the proof. (I do not know whether this can be proved for arbitrary normed space, i.e., whether we can omit this assumption.)
The equality you are asking about is a special case of this, since you are asking basically about two partitions of $K\times \Lambda$. But notice that one of the assumption of this theorem is that the sum $\sum\limits_{(\kappa,\lambda)\in K\times\Lambda} a_{\kappa,\lambda}$ exists. Under this assumption it says that either of the two sums will give the same result.
I will add that the definition of $\sum\limits_{\lambda\in\Lambda} a_\lambda$ using nets is given, for example, in
this answer. The definition using supremum of finite sums (which is equivalent in the case of non-negative reals) is mentioned in several answers to
this question.
