Q:

There appears to be two definitions of the direct sum of vector spaces

I have seen two definitions of the direct sum of vector spaces:

If $U$ and $V$ are vector spaces, then

$$ U \oplus V = \{(u,v):u\in U,v \in V\}.$$

If $V_1$ and $V_2$ are subspaces of $V$, and $V_1 \cap V_2 = \{0\} $, then

$$ V_1 \oplus V_2 = \{u+v:u\in V_1,v \in V_2 \}$$
So essentially, if I was to take the direct sum of the vector spaces $V_1$ and $V_2$ consisting of vectors of the form $(a,b,0,0)$ and $(0,0,c,d)$ respectively, would it be as space with vectors of the form
$$ (a,b,0,0,0,0,c,d) $$
by definition 1, or
$$ (a,b,c,d) $$
by definition 2?

A:

The two notions are the same (up to isomorphism) for vector spaces whose intersection is $\{0\}$. Assume $U$ and $V$ are vector spaces such that $U \cap V = \{0\}$. I'll refer to their external direct sum $(1)$ as $U\times V$ and their internal direct sum $(2)$ as $U \oplus V$.
As mentioned in the comments $U \times V \cong U \oplus V$, and the isomorphism is the natural choice. The only thing that requires a bit of thinking is noticing where the condition $U\cap V = \{0\}$ comes in. The isomorphism is the following:
$$
\begin{align}
\phi : U \times V &\to U \oplus V \\ (u,v) &\mapsto u + v
\end{align}
$$
It is straightforward to check that this is a homomorphism of vector spaces (a linear map). In addition you need to check that it is bijective. It is immediate that $\phi$ is surjective. To see that it is injective, note that $\ker(\phi) = \{(u,v)\mid u + v = 0\}$. But $u + v = 0 \implies u = -v$. This means that $-v \in U$ and since $U$ is a vector space this implies $v \in U$. By the assumption $U \cap V = \{0\}$ we have that $v = u = 0$. So $\ker(\phi) = \{0\}$, proving that $\phi$ is injective.  

