As in the question, we set $n := \dim V$ and $m := \dim W$ and assume both are finite (though many conclusions hold just as well when either or both are infinite), and pick arbitrary bases $A = (v_a)$ of $V$ and $B = (w_b)$ of $W$. Also, we usually regard bases as ordered sets of vectors; the below doesn't always specify orders, but of course these can be chosen arbitrarily to produce a bona fide basis.
Sums $V + W$ Pick a basis $(x_c)$ of $V \cap W$ (denote $p := \dim (V \cap W) = p$), and extend it to a basis $(x_1, \ldots, x_p, y_1, \ldots, y_{m - p})$ of $W$. Then, $$(v_1, \ldots, v_n, y_1, \ldots, y_{m - p})$$ is a basis of $V + W$, which thus has dimension $n + m - p$; usually this is written $$\dim(V + W) = \dim V + \dim W - \dim (V \cap W).$$
Direct sums $V \oplus W$ Via some identifications we can view this as a special case of (1), but we may as well be explicit: $$(A \times \{0\}) \cup (\{0\} \times B) = \{(v_a, 0) : v_a \in A\} \cup \{(0, w_b) : w_b \in B\}$$ is a basis of $V \oplus W$, and so $$\dim (V \oplus W) = \dim V + \dim W.$$ For convenience, we often mildly abusively write $(v_a, 0)$ as $v_a$ and $(0, w_a)$ as $w_a$, in which case we can write the basis as
$$A \cup B = (v_1, \ldots, v_n, w_1, \ldots, w_m).$$
Cartesian products $V \times W$ Since the product has a finite number of factors, this is indeed the same as $V \oplus W$ in (2); see The direct sum $\oplus$ versus the cartesian product $\times$ for more.
Tensor products $V \otimes W$ $$\{v_a \otimes w_b : v_a \in A, w_b \in B\} \cong A \times B$$ is a basis for $V \otimes W$ and hence $$\dim (V \otimes W) = \dim V \cdot \dim W.$$
Multivector spaces $\bigwedge^k V$
$$\{v_{a_1} \wedge \cdots \wedge v_{a_k} : 1 \leq a_1 < \cdots < a_k \leq n\},$$ and so $$\dim \bigwedge^k V = {{\dim V}\choose{k}}.$$ By convention, $\bigwedge^0 V$ is just the field $\mathbb{F}$ underlying $V$, and we take the empty wedge product to be $1 \in \mathbb{F}$, so $(1)$ is a basis of $\bigwedge^0 V$; $\bigwedge^k V$ is trivial for $k > m$, in which case the given basis is empty.
Symmetric powers $\bigodot^k V$
$$\{v_{a_1} \odot \cdots \odot v_{a_k} : 1 \leq a_1 \leq \cdots \leq a_k \leq n\}$$ is a basis for $\bigodot^k V$, where $\odot$ defines the symmetric product, and one can then show $$\dim \bigodot^k V = {{\dim V + k - 1}\choose{k}}.$$
For completeness we should also mention one more:
Tensor powers $\bigotimes^k V$
$$\{v_{a_1} \otimes \cdots \otimes v_{a_k} : v_{a_1}, \ldots, v_{a_k} \in A\} \cong A \times \cdots \times A$$ is a basis for $\bigotimes^k V$, so that $$\dim \bigotimes^k V = (\dim V)^k.$$
Bivector space over a direct sum Finally, we can treat $\bigwedge^2 (V \oplus W)$. By (2), a basis for $V \oplus W$ is $$A \cup B = \{v_1, \ldots, v_n, w_1, \ldots, w_m\},$$
and then by (5), a basis of $\bigwedge^2 (V \oplus W)$ is
\begin{multline}
(v_1 \wedge v_2, \ldots, v_1 \wedge v_n, v_1 \wedge w_1, \ldots, v_1 \wedge w_m, v_2 \wedge v_3, \ldots, w_{m - 1} \wedge w_m) \\
= \{v_a \wedge v_{a'} : a < a'\} \cup \{v_a \wedge w_b\} \cup \{w_b \wedge w_{b'} : b < b'\}.
\end{multline}
The elements $v_a \wedge v_{a'}$ comprise a basis for $\bigwedge^2 V$, and likewise those of the form $w_b \wedge w_{b'}$ comprise a basis for $\bigwedge^2 W$. The forms $v_a \wedge w_b$ span a space that can be identified with $V \otimes W$ via the map characterized by the isomorphism $v_a \otimes w_b \mapsto v_a \wedge w_b$. This gives a natural decomposition $$\bigwedge^2 (V \oplus W) \cong \bigwedge^2 V \oplus (V \otimes W) \oplus \bigwedge^2 W.$$ Of course, the dimension formulas produced above lead to the same result for the dimension of $\bigwedge^2 (V \oplus W)$: $${{n + m}\choose 2} = \frac{1}{2}(n + m)(n + m - 1) = \frac{1}{2}n(n - 1) + nm + \frac{1}{2}m(m - 1) = {n \choose 2} + nm + {m \choose 2}.$$
