Direct product vs direct sum of infinite dimensional vector spaces?
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.
$V\times W$ and $V\oplus W$ are isomorphic, as are any finite sums/products of spaces. This is true for any category of modules.
When $I$ is infinite $ \bigoplus_{i \in I} V_i $ and $ \times_{i \in I} V_i$ can be very different. For example, when $V_i=F_2$ the field of two elements, and $I=\mathbb N$, then the left hand one is countable while the right hand one is uncountable. So they obviously cannot be isomorphic. I think this is probably always the case by some cardinal arithmetic, but at this very moment I can't definitively say.
The definition of finite direct sum and the definition of finite direct product is exactly the same definition. (Unless you are working in categories and then the definitions aren't the same, but as vector spaces they are isomorphic).
Infinite direct sum of $\{V_i\}_{i\in I}$ is the set of all finite sums of vectors, while the direct product is as a set, the product of all $V_i$, so, in particular, the direct sum does not contain the element $(1,1,1,1,....)$ while the direct product does.
For your questions
1)V and W are infinite dimensional, but since you only take the direct sum/product once the definitions are equal.
2) No, because the element (1,1,1,1,...) is not an element in the direct sum, but it is an element in the direct product.