The writing
means that $V_1$ and $V_2$ are subspaces of $V$, such that every vector $v$ in $V$ can be written, in one and only one way, as a sum
$$v=v_1+v_2$$
where $v_1\in V_1$ and $v_2\in V_2$. The unicity part of this statement is equivalent to the fact that $V_1$ and $V_2$ trivially intersect
$$V_1\cap V_2=\{\vec 0\}$$
Another way to characterize the direct sum of two subspaces could be that $V_1\oplus V_2$ is the subspace generated by the union $V_1$ and $V_2$, meaning the smallest subspace of $V$ containing both vectors of $V_1$ and $V_2$
There is no difference between the direct sum and the direct product for finitely many terms, regardless of whether the terms themselves are infinite-dimensional or not. However, they are different in the case of infinitely many terms (and drastically so).
A direct product $\prod_{i = 1}^\infty V_i$ can be thought of as the set of sequences $(v_1, v_2, \ldots)$ where each $v_i \in V_i$, with usual scalar multiplication $\lambda (v_1, v_2, \ldots) = (\lambda v_1, \lambda v_2, \ldots)$ and pointwise addition $(v_1, v_2, \ldots) + (w_1, w_2, \ldots) = (v_1 + w_1, v_2 + w_2, \ldots)$. The direct sum $\bigoplus_{n=1}^\infty V_i$ on the other hand is the same set, with the extra condition that only finitely many terms are nonzero.
The direct sum behaves nicely in terms of bases. If each $V_i$ has some basis set $B_i \subseteq V_i$, then the direct sum $\bigoplus_{n=1}^\infty V_i$ has a basis identified with $\bigsqcup_{i=1}^\infty B_i$. For example, if all $V_i = \mathbb{R}$, then the basis for the direct product is just putting a 1 in the $i$th place for all $i$: $(1, 0, 0, \ldots), (0, 1, 0, 0, \ldots), \ldots$. In particular, if all the $V_i$ are countable dimension, the direct sum is also countable dimension.
With the direct product, this is not the case. It is not so hard to see (I'm sure there are many answers on this site) that the space of sequences of real numbers $\prod_{i=1}^\infty \mathbb{R}$ has uncountable dimension over $\mathbb{R}$.
Finally, there is not that much subtlety in what it means to be a basis of an infinite dimensional space. A basis is a linearly independent subset $B \subseteq V$ such that any vector $v \in V$ may be written as a finite linear combination of basis vectors. This is perhaps the best way to think about the difference between direct sum and product: start trying to write down a system of elements which can express any real sequence as a finite linear combination, and you'll soon see that in many cases, a direct sum may have been what you intended all along.
Best Answer
There are three standard ways of combining a collection of vector spaces $V_i$, and in full generality they are all different:
The direct sum and direct product agree for finitely many factors but disagree in general; the tensor product almost never agrees with either.
So much for vector spaces; what about matrices? The universal properties of the direct sum and direct product can concisely be written as
$$\text{Hom}(\bigoplus V_i, W) \cong \prod \text{Hom}(V_i, W)$$
and
$$\text{Hom}(W, \prod V_i) \cong \prod \text{Hom}(W, V_i).$$
It follows that
$$\text{Hom}(\bigoplus V_i, \prod W_i) \cong \prod \text{Hom}(V_i, W_j).$$
So given a collection of maps $f_{ij} : V_i \to W_j$ we canonically get a map $f : \bigoplus V_i \to \prod W_i$. For finitely many factors we have $\prod W_i \cong \bigoplus W_i$, and so in this case I guess one could call $f$ the "direct sum" of the $f_{ij}$, although I think this is mildly misleading. I don't know a better term, though.
What I've described above is not the Kronecker product. The Kronecker product is a description in coordinates of an abstract way to combine a collection of maps $f_i : V_i \to W_i$ into a map $f : \bigotimes V_i \to \bigotimes W_i$, as follows: given a multilinear map $B$ from the $W_i$ to some vector space $U$, we can compose $B$ with each of the $f_i$ to get a multilinear map from the $V_i$ to $U$, and by the universal property we get the desired map $f$.
The Kronecker product is entirely defined in terms of the tensor product, and in particular makes no use of the direct product, so I think it is quite misleading to call it the "direct product."