"Internal" and "external" in maths, and also in vector spaces I have looked at 3 books and it is clear that "internal" and "external" are two styles of defining something, I would like to know what they mean "generally" - that is very soft but it is clear to me that "internally" doing something is different from doing it "externally" even if the result is similar.
I've read ahead and motivating examples or even uses of these definitions are not given, so if you have any examples that come to mind please do share. 
First it defines External direct sum using a $+$ inside a $\square$ of a finite number of vector spaces over the same field (I understand "external" might mean "not considering them as subspaces of something")
$V=V_1+...+V_n$ is defined component wise (as vectors themselves)
This is fine, I am happy with this.
Direct product next and we have a family $\funky{F}=\{v_i|i\in K\}$ the direct product is:
$\Pi_{i\in K}V_i=\{f:k\rightarrow\cup_{i\in K}V_i|f(i)\in V_i\}$ 
At first this seems horrible but really it's not too bad, $f$ is acting as a projection to coordinates really. This is a vector space itself and thinking about it I ask myself "How is that not the external direct sum (provided the family is finite)"
This is not what I was taught for sum take the plane $x=0$ and the plane $y=0$ in $\mathbb{R}^3$, their union is just an extruded $+$ shape but what I was taught was the sum becomes all of $\mathbb{R}^3$ because the sum was defined as $\{u+v|u\in V_1 v \in V_2\}$
But anyway, that's direct product.
External direct sum
This time it is denoted not by a $+$ in a square but by a + in a circle with "ext" in superscript, due to lack of knowledge of the name of this symbol I must describe it to you.
Anyway this sum is (on a family as above) $\{f:K\rightarrow\cup_{i\in K}V_i|f(i)\in V_i, f\text{ has finite support}\}$
finite support means $f(i)=0$ for all but a finite number of $i$
I dare not come up with examples because it's not finite. 
Lastly Internal direct sums
This is written as a + with a circle around it, and it requires the following hold:
The join of the family is V, that is: $V=\sum_{i\in K}S_i$ which is the more conventional sum I assume, and independence of the family, that is:
$S_i\cap(\sum_{j\ne i}S_j)=\{0\}$
It notes that this second condition is stronger than pairwise disjoint, I cannot think of an example which shows this distinction and I'd like one.
That's all the book does on these, unless they are used far later. What is going on here? Also what do "internal" and "external" mean, I can't think of an example but I think it refers to a style of definition, I've come across and gotten used to similar things before.
The book is: Advanced Linear Algebra, Steven Roman, 3rd Edition, GTM # 135
 A: The question is quite interesting, at least from an epistemological point of view. It is not easy to work out an all-comprehensive answer which is in the meantime meaningful enough in a broad context, but I will try to say something that hopefully can be understood, at least on a metamathematical level, whatever this may mean. Also, since you have tagged the question under "Category Theory", I will feel free to use that language, which, unsurprisingly, provides quite an enlightening way of interpreting the problem.
So, the main difference between an "internal" and an "external" approach may be summarised as follows. Suppose you have fixed a mathematical entity (or structure) $\mathcal{M}$ which you are interested in studying, for example a vector space $V$. For our purposes, I guess one should think to such a mathematical entity as a mathematically definable object, characterised by some defining properties (of course, this "definition" is deliberately vague).
Investigating the nature of $\mathcal{M}$ internally means to study $\mathcal{M}$ using solely the structure that $\mathcal{M}$ has in its own and the properties that the axioms defining $\mathcal{M}$ imply, as if $\mathcal{M}$ had suddenly become the only available working ground, a comprehensive universe into which doing your math. In particular, one is not allowed to refer to any other entity besides $\mathcal{M}$ or besides those that can be defined or built up in $\mathcal{M}$. For example, if $\mathcal{M}$ is your favourite vector space $V$, looking at its linear subspaces is an internal way of proceeding, because subspaces are substructures of your given entity. You can then do some operations on subspaces, such as taking intersections or internal direct sums, which are internal as long as they can be performed inside $V$ and their result still lies inside $V$.
On the other hand, an external study of $\mathcal{M}$ would try to grasp information about $\mathcal{M}$ basing on the connections that $\mathcal{M}$ may have with other entities, or, better, seeing $\mathcal{M}$ as within (or as related to) a wider landscape, a working universe to which one is allowed to link $\mathcal{M}$. For instance, once you have taken your favourite vector space $V$ and a bunch of subspaces $V_{i}$ of $V$, you may realise that such subspaces are themselves vector spaces and constitute mathematical entities on their own right, even if of a kind similar to the one of your $V$. In other words, both $V$ and the $V_{i}$ are objects of the category of vector spaces (over some field). Then, you may perform some operations on these $V_{i}$ seeing them as structures on their own, and not as substructures or $V$ (i.e. considering them as objects in the category of vector spaces and not as subobjects of $V$). For example, you may take the external direct sum, or the direct product of the $V_{i}$ and then, you may wonder if the results of such operations can be related to your $V$, giving some informations on it ($V$ may be isomorphic to the external direct sum of the $V_{i}$ or a quotient of them).
The best way to explain the two approaches comes thinking at Category Theory. Take a (locally small) category $\mathcal{C}$ (even better, an elementary topos). Then such a category is naturally related to the category $\mathbf{Set}$ of sets through the Hom functors. Now, one can study properties of $\mathcal{C}$ or even define some notions internally to $\mathcal{C}$, using only the given structure of $\mathcal{C}$, i.e. its objects, its arrows and its composition law, or externally transferring the properties and the definitions to the category of sets.
For instance, given objects $X,Y$ of $\mathcal{C}$, one can define their product $X\times Y$ either internally, requiring that $X\times Y$ is an object of $\mathcal{C}$ equipped with two morphisms in $\mathcal{C}$ satisfying the well-known universal property, or externally, requiring that the functor
$$
\mathcal{C}(-, X)\times \mathcal{C}(-,Y)
$$
is representable, i.e. requiring that, for all $Z\in\mathcal{C}$, $\mathcal{C}(Z,X\times Y)$ is a product in the category of sets. The same kind of double approach can be carried over to loads of notions commonly used in Category Theory, such as limits, colimits, subobject classifiers, exponentials etc. 
It turns out that (at least in the abovementioned examples) the two approaches are equivalent, in the sense that an object (together with a family of arrows) in $\mathcal{C}$ satisfy a property internally if and only if it satisfies it externally.
However, the two ways of investigating the category $\mathcal{C}$ are sensibly different: the internal one makes assertions just about the structure of $\mathcal{C}$, without any further (set-theoretic) assumptions and can therefore be used to give a self-referring description of $\mathcal{C}$, as if it was the universe into which we set our play. On the other hand, the external point of view studies $\mathcal{C}$ as a structure constructed within a working, fundational environment (some kind of set-theory), into which (and dipendently to which) one develops the whole theory on $\mathcal{C}$.
The internal and the external point of view are a common feature in Category Theory (hence, in the whole Mathematics) and their interplay and connections are not only what makes the subject delightful but provide deep results (just to mention an example that springs to my mind, think about Giraud's Theorem characterizing axiomately a Grothendieck topos).
A: The distinction between internal and external direct sums of vector spaces (or of groups, or of other algebraic gadgets) is a more complicated instance of the fact that not all unions of subsets are disjoint. 

Recall that a disjoint union $\bigsqcup_{i\in I}X_i$ of a family of sets $(X_i)_{i\in I}$ (so indexed by the set $I$) can be constructed as the set $\bigsqcup_{i\in I}X_i=\{(i,x):i\in I,x\in X_i\}$, plus the data of the obvious functions $X_i\to\bigsqcup_{i\in I}X_i$ given by $x\mapsto (i,x)$.
On the other hand, the union $\bigcup_{i\in I}X_i$ of a family of subsets $X_i\subseteq X$ is defined to be the subset $\{x\in X:\exists i\in I\,(x\in X_i)\}\subseteq X$ of $X$; implictily the union of subsets also has the data of maps $X_i\to\bigcup_{i\in I}X_i$, which are given by $x\mapsto x$. 
Now, the disjoint union of a family of sets $X_i$ has a univeral property: given any family of functions $X_i\overset{f_i}\rightarrow Y$ into some set $Y$, there is a unique function $\bigsqcup_{i\in I}X_i\overset{f}\dashrightarrow Y$ so that each $f_i$ is the composite of $f$ with the injective map $X_i\to\bigsqcup_{i\in I}X_i$. Indeed, the condition on the composite says that given $x\in X_i$, its image $f_i(x)$ under $X\overset{f_i}\to Y$ must be the image $f((i,x))$ of the image $(i,x)\in\bigsqcup_{i\in I}X_i$ of $x\in X_i$ under the injective map $X_i\to\bigsqcup_{i\in I}X_i$. But this determines uniquely the function $\bigsqcup_{i\in I}X_i\overset f\to Y$ as the one sending $(i,x)\mapsto f_i(x)$.
(The significance of universal properties is that any two objects satisfying the same universal property have a natural isomorphism between them. This allows to make rigorous the fact that some construction seem to involve choices that are arbitrary matter only insofar as us having to make them in order to actually carry out the construction. For example, we could have just as easily constructed a disjoint union as $\{(x,i):x\in X_i\}$ instead of $\{(i,x):x\in X_i\}$. These two sets are different, and which one we use as the disjoint union is a completely arbitrary choice, but we can use the fact that both of them satisfy the universal property of disjoint union to construct the natural bijection (i.e. isomorphism) sending $(x,i)\mapsto(i,x)$.)
Next, since the union $\bigcup_{i\in I}X_i$ comes with maps $X_i\to\bigcup_{i\in I}X_i$, there is a natural map $\bigsqcup_{i\in I}X_i\to\bigcup_{i\in I}X_i$ from the disjoint union of the subsets (considered as sets) to their union as subsets. This natural map simply sends $(i,x)\mapsto x$. You can easily see that the map is surjective. In the case where the map is also injective, then the disjoint union is isomorphic to the union, so we say that the union is disjoint.
The above relationship between the disjoint union and union comes about from the fact that the union satisfies a weaker universal property: namely, it is for subsets $Y\subseteq X$, not arbitrary sets $Y$, that there is a unique function $\bigcup_{i\in I}X_i\overset f\to Y$ given a family of functions $X_i\xrightarrow{f_i}Y$ so that each $f_i$ is the composite of $f$ with $X_i\to\bigcup_{i\in I}X_i$.
Note finally that in order for the union of subsets $X_i\subseteq X$ to be disjoint, it is necessary that the family of subsets be pairwise disjoint, i.e.\ that $X_i\cap X_j=\emptyset$ if $i\neq j$. Indeed, this is the property of the union of the isomorphic subsets $\{(i,x):x\in X_i\}\subseteq \bigsqcup_{i\in I}X_i$, hence a union isomorphic to the disjoint union must also have the property. It turns out this property of being pairwise disjoint is also sufficient (for sets anyway, as we'll see below for vector spaces it is not sufficient).

Working with vector spaces, the external direct sum $\bigoplus^{ext}_{i\in I}V_i=\{$sequencs $(v_i\in V_i)_{i\in I}$ so that $v_i=0$ for all but finitely many $i\in I\}$ comes with obvious linear maps $V_i\to\bigoplus^{ext}_{i\in I}V_i$ given by $v\mapsto(v_j)_{j\in I}$ where $\begin{cases}v_j=v&j=i\\0&\text{otherwise}\end{cases}$ and has the same universal property as the disjoint union, but for linear maps instead. Indeed, given any family of linear maps $V_i\overset{f_i}\to W$, we have that $\bigoplus^{ext}_{i\in I}V_i\overset{f}\dashrightarrow W$ must be given by $f((v_i))=\sum f_i(v_i)\in W$ which is a finite sum by the finite support property.
On the other hand, the sum $\sum_{i\in I}V_i$ of a family of vector subspaces $V_i\subseteq V$ is given by $\sum_{i\in I}V_i=\{v\in V:v=$a finite sum $\sum v_i$ with $v_i\in V_i\}$. It also has natural linear maps $V_i\to \sum_{i\in I}V_i$ given by $v\mapsto v$, and satisfies same universal property as union of subsets, but with subspaces rather than subsets. Explicitly, it is for subspaces $W\subseteq V$ that there is a unique linear map $\sum_{i\in I}V_i\overset f\to W$ given a family of functions $V_i\xrightarrow{f_i}Y$ so that each $f_i$ is the composite of $f$ with $V_i\to\sum_{i\in I}V_i$.
Using the universal properties, we see that the external direct sums relate to sums the same way that disjoint unions relate to unions. In particular, there is always a natural map $\bigoplus^{ext}_{i\in I}V_i\to\sum_{i\in I}V_i$ sending $(v_i)_{i\in I}\mapsto\sum v_i$. We say that a sum of subspaces $V_i\subseteq V$ is an internal direct sum if the natural map $\bigoplus^{ext}_{i\in I}V_i\to\sum_{i\in I}V_i$ is an isomorphism. In this case we denote the sum $\sum_{i\in I}V_i$ of subspaces by $\bigoplus_{i\in I}V_i$.
The reasoning behind this choice of notation is this: for vector spaces, it is always that case that the natural maps $V_i\to\bigoplus^{ext}_{i\in I}V_i$ are injective linear maps, hence that an external direct sum is also an internal direct sum.
Once again, the external direct sum has the property that the isomorphic subspaces $V_i\hookrightarrow\bigoplus_{i\in I}^{ext}V_i$ of the external direct sum considered as an internal direct sum, are pairwise disjoint (in the sense that their pairwise intersections are the zero vector subspace). However, unlike the case of unions of subsets, this is not a sufficient condition for the sum of vector subspaces to be the analogue of disjoint, i.e. an internal direct sum.
Indeed, consider $\mathbb R^2$ and the subspaces given by the $x$-axis, the $y$-axis, and a diagonal axis. The three axes are certainly pairwise disjoint, yet their sum is $\mathbb R^2$, while their disjoint sum is isomorphic to $\mathbb R^3$. It follows that the sum of these three pairwise disjoint subspaces is not an internal direct sum.
It turns out (and it is a good exercise I think) that the necessary and sufficient condition is $V_i\cap\sum_{j\neq i}V_j)=\{0\}$, which Roman takes as the definition of an internal direct sum.
