There are structures that are defined as various direct sums, like tensor algebras $T(V)$, exterior algebras $\Lambda(V)$ and so on. While all these structures are supposed to be external direct sums, which are cartesian products, they are almost always treated as internal ones, which are sums of subspaces.
And in each case summands are treated as subsets. While I can see how this works in case of internal direct sums, but it is clearly that for external case $C=A\oplus_E B$, neither $A$ nor $B$ are the subsets of $C$. It is more likely that $A\times\{0\}\subset C$ and $\{0\}\times B \subset C$, s.t. $(A\times\{0\})\oplus_I(\{0\}\times B)=C$
It confuses me alot, because it is often that operations defined on external sums, like wedge product, are used on summands though they have different underlying sets. Of course one can construct operation that will works on each summand from the given on the whole space, but that would be already different operation and we would need to distinguish between them.
Could you please clarify to me how this works?
I think that operations like direct sum are best understood first as internal operations. Let $R$ be some ring, with $M$ a left $R$-module. If I can write $M = N \oplus P$ as $R$-modules, this means that I have found a decomposition of $M$ such that $R$ will act on each component in an isolated way. In particular, any $m \in M$ may be written as $m = n + p$ for $n\in N$, $p \in P$, and then we have $rm = rn + rp$, where $rn \in N$ and $rp \in N$. Crucially, this composition does not "mix" under the action of $R$.
The external direct sum is then understood as a method of taking the modules $N$ and $P$ and constructing a module with the same behaviour of $M$ with respect to this decomposition, which symbolically just means replacing the $+$ sign in the above with a comma: $x = (n, p)$ and then $rx = (rn, rp)$. There is an obvious homomorphism from $N \oplus_E P \to M$, just be doing $(n, p) \mapsto n + p$, which is an isomorphism. So we may say that the module we cooked up is isomorphic to $M$, and so to some degree it doesn't matter whether we consider the direct sum $N \oplus P$ as being external, or sitting inside $M$.
Sometimes it may be easier to regard $M$ on its own, and other times it may be useful to extract factors out of the decomposition $M = N \oplus P$. If we want to go between these modes of thought a lot (and sweep the gritty details under the rug), we need an easy way of telling how an element decomposes into $N$ and $P$. For your examples, the tensor algebra and the exterior algebra have an obvious grading, being the rank of the tensor. If I write "The submodule of $T(V)$ generated by all $v \otimes v$", its obvious that this lives in the degree 2 part of the grading, which would be the $i = 2$ part of the exterior direct sum $$T(V) = \bigoplus_{i \geq 0} V^{\otimes i}$$ so most people will be happy going between internal and external direct sums based on the grading.