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.
The core idea is that if you have two different ways to decompose the same vector
$$
u_1 + \cdots + u_m = v_1 + \cdots + v_m
$$
(where $u_i, v_i\in U_i$) then the zero vector can be decomposed as
$$
0 = (u_1 - v_1) + (u_2 -v_2) + \cdots + (u_m - v_m)
$$
and since the two original decompositions were different, not all these terms can cancel, and we have found a second way to decompose the $0$ vector.
So if there is some vector which can be decomposed in two different ways, then the $0$ vector can be decomposed in two different ways. Contrapositively, if the $0$ vector can only be decomposed as $0 = u_1 +\cdots + u_m$ with $u_1 = \cdots = u_m = 0$, then any other vector also only has one deomposition.
Best Answer
Yes. If $V$ is any vector space and $(W_i)_{i\in I}$ is a family of subspaces of $V$, we write $$V =\bigoplus_{i\in I}W_i$$if the following conditions hold:
(i) for every $v\in V$ there are $i_1,\ldots,i_k\in I$ ($k$ depends on $v$) and $w_{i_1}\in W_{i_1}$, ..., $w_{i_k}\in W_{i_k}$ such that $$v=w_{i_1}+\cdots + w_{i_k}.$$This is written as $V=\sum_{i\in I}W_i$.
(ii) for every $j\in I$, we have $W_j\cap \sum_{i\neq j}W_i = \{0\}$.
Condition (ii) ensures that the decomposition of $v$ in (i) is unique. For example, if $(e_i)_{i\in I}$ is a basis for $V$ over a field $\mathbb{K}$, then $$V=\bigoplus_{i\in I}\mathbb{K}e_i,$$where each $\mathbb{K}e_i$ is the line spanned by $e_i$.