When I learned linear algebra I was a bit confused about the notation $A \oplus B$, the direct sum of two vector spaces $A, B$ over some field $K$. As a set, $A \oplus B$ is nothing else than the Cartesian product $A \times B$ and the vector space operations are defined componentwise. So why not write $A \times B$ and call it the direct product of $A, B$? My lecturer said that one can do this and that the difference between both concepts becomes visible when we consider infinite families $A_i$, $i \in I$, of vector spaces. Then the direct product $\prod_{i \in I} A_i$ is the Cartesian product of the $A_i$ having as elements all "tuples" $(a_i)$ with $a_i \in A_i$ and the vector space operations are defined componentwise. In contrast the direct sum $\bigoplus_{i \in I} A_i$ is the linear subspace of $\prod_{i \in I} A_i$ having as elements all tupels $(a_i) \in \prod_{i \in I} A_i$ such that $a_i \ne 0$ at most for finitely many $i \in I$.
After having learned category theory, I realized that this gives an incomplete picture. Direct product (= categorical product) and direct sum (= categorical coproduct aka categorical sum) are more than just objects, but come along with certain morphisms ("projections" $\pi_i : \prod_{i \in I} A_i \to A_i$ and "inclusions" $\iota_i : A_i \to \bigoplus_{i \in I} A_i$) such that certain universal properties are satisfied.
That it is possible to take the same object for $A \times B$ and $A \oplus B$ is a special property of a category, and the category of vector spaces over a field $K$ has this property.
A good motivation for the use of the phrase direct sum is the internal direct sum of subspaces $A, B$ of a vector space $V$. For any two subspaces $A, B$ we can define $$A + B = \{ a + b \mid a \in A, b \in B \}$$ which is always a subspace of $V$. If $A \cap B = \{0\}$, then $A + B$ is called the internal direct sum of $A, B$. In that case each $w \in A + B$ has a unique representation $w = a + b$ with $a \in A, b \in B$. In other words, the map $$\phi : A \oplus B \to A + B, \phi(a, b) = a + b$$ is an isomorphism of vector spaces.
I have not seen that the internal direct sum is defined for a family $A_i$, $i \in I$, of subspaces $A_i$ of $V$ with more than two members, but it can be done. For any such family we can define $$\sum_{i \in I} A_i = \{\sum_{i \in I} a_i \mid a_i \in A_i, a_i \ne 0 \text{ at most for finitely many } \in I \} .$$
To obtain an internal direct sum, it does not suffice to require $A_i \cap A_j = \{0\}$ for $i \ne j$. The adequate condition is that
For any finite $J \subset I$ and any family of $a_j \in A_j \setminus \{0\}$ the equation $\sum_{j \in J} \lambda_j a_j = 0$ implies that all $\lambda_j = 0$.
Of course this condition implies $A_i \cap A_j = \{0\}$ for $i \ne j$. In case $I = \{1,2\}$ it is easy to see that the condition is equivalent to $A_1 \cap A_2 =\{0\}$.
Under the above condition we obtain an isomorphism of vector spaces $$\phi : \bigoplus_{i \in I} A_i \to \sum_{i \in I} A_i, \phi((a_i)) = \sum_{i \in I} a_i .$$
Here is my question:
Is there a concept of an internal direct product of subspaces? This is relevant only for infinite families of subspaces.
This answer is somewhat of a frame challenge.
Subspaces by definition come with (injective) maps to the ambient space, $A \to V$ and $B \to V$. Therefore by the universal property you get a canonical map $A \oplus B \to V$, which has kernel the fiber product or pullback $A \times_V B$ which (for two terms) you can identify with $A \cap B$. So the map is injective iff $A \cap B = 0$, in which case it is an isomorphism onto its image, which is called the internal direct sum. Once you modify the $A \cap B = 0$ criterion in the way you describe (or ignore it) this definition generalizes to more than two summands.
If you dualize everything in this picture you should consider internal direct product of quotients. Given maps $V \to A_i$, there is a canonical map $V \to \prod A_i$, and I would call this an internal direct product if it is surjective. This necessarily means that each $V \to A_i$ is surjective but that is not sufficient.