I'm interested in understanding compositum of general fields better. Assume we have $\Omega/K/F$ and $\Omega/L/F$ field extensions, and consider the composite $KL$.
It seems to me that every $a \in KL$, exists $n_1,n_2 \in \mathbb{N}$, $ \left \lbrace k_i \right \rbrace_{i=1}^{n_1}, \left \lbrace k'_i \right \rbrace_{i=1}^{n_2} \subseteq K$ and $ \left \lbrace l_i \right \rbrace_{i=1}^{n_1}, \left \lbrace l'_i \right \rbrace_{i=1}^{n_2} \subseteq L$ such that: $$ a = \frac{\sum_{i=1}^{n_1} k_i l_i}{\sum_{j=1}^{n_2} k'_j l'_j} $$
(That is since the family of all such sums is closed under addition, multiplication, and distribution because of the abelian nature of fields, therefore all numbers of this form are subfields that contain both K and L)
Is it a correct way to refer to a compositum of fields? When is it correct that the general item of an extension is $a = \sum_{i=1}^{n} k_i l_i$?
You are correct. In general, if $S$ is a subset of $\Omega$ and $F$ is a subfield, then $F(S)$ is defined to be the intersection of all subfields of $\Omega$ which contain $F$ and $S$. You can show that $F(S)$ is the set of all quotients $\frac{f(s_1, ... , s_n)}{g(s_1, ... , s_n)}$, where $n \geq 1, s_i \in S$, and $f, g \in F[X_1, ... , X_n]$ with $g(s_1, ... , s_n)$.
By definition, $KL$ is the field $K(L)$. But you can show this is the same thing as $L(K)$. In fact, $$KL = K(L) = L(K) = F(K \cup L)$$ (you can replace $F$ by any subfield of $\Omega$ which is contained in $L \cap F$) and this last field $F(K \cup L)$ you can easily show to be the same as the thing you said. I am not sure if any of these descriptions is simpler than another.
Again if $F$ is a subfield of $\Omega$ and $S$ a subset, $F[S]$ is defined to be the intersection of all subrings of $\Omega$ containing $F$ and $S$. You can show that $F[S]$ is equal to the set of all $f(s_1, ... , s_n)$, where $n \geq 1, s_i \in S$, and $f \in F[X_1, ... , X_n]$. Again $$K[L] = L[K] = F[K \cup L]$$
Here is a useful proposition:
Proposition: The following are equivalent:
(i): $F(S) = F[S]$.
(ii): Every element of $S$ is algebraic over $F$.
Proof: Let me think about it. It should follow from the case when $|S| = 1$, which is a pretty standard result.
From the proposition, we can make the following observation, $K[L]$ is equal to the set of all finite sums $k_1l_1 + \cdots + k_tl_t, k_i \in K, l_i \in L$. So we see that $KL$ admits the nice description you suggested if and only if one of the following equivalent conditions are satisfied:
(i): $KL$ is algebraic over $K$.
(ii): $KL$ is algebraic over $L$.
(iii): $KL$ is algebraic over $K \cap L$.
(iv): Every element of $L$ which is not in $K$, is algebraic over $K$.
(v): Every element of $K$ which is not in $L$, is algebraic over $L$.