Let $A$ be a commutative ring. Then $\Lambda(-)$ is a functor from $A$-modules to graded-commutative $A$-algebras which is left adjoint to the functor which takes the degree $1$ part. Because it is left adjoint, it preserves colimits, in particular coproducts. It follows $\Lambda(M \oplus N) \cong \Lambda(M) \otimes \Lambda(N)$. Looking at the $n$th degree part, we obtain $\Lambda^n(M \oplus N) \cong \bigoplus_{p+q=n} \Lambda^p(M) \otimes \Lambda^q(N)$.
For a more direct proof, consider the left hand side as a quotient of $(M \oplus N)^{\otimes n}$ and the right hand side as a quotient of $\bigoplus_{p+q=n} M^{\otimes p} \otimes N^{\otimes q}$. We have the "binomial theorem" $(M \oplus N)^{\otimes n} \cong \bigoplus_{p+q=n} \binom{n}{p} \cdot M^{\otimes p} \otimes N^{\otimes q}$. One easily checks that the quotients agree.
Both arguments work in great generality, $A$ can be any commutative monoid object in a cocomplete linear symmetric monoidal category (this is explained, for example, in my thesis). As usual, there is no need to fiddle around with elements.
Your reasoning is not necessarily incorrect, but the issue you see can be mitigated. To do this, we have to slightly ponder what "kind of object" exterior algebras are and what the symbol $\otimes$ means. Namely, if $V$ is a $k$-vector space, then $\bigwedge V$ is not just a $k$-algebra, but a graded $k$-algebra. In fact, it is a graded commutative $k$-algebra (the Wikipedia article calls this "anticommutative" instead - I personally disagree with that choice of terminology), i.e. if $\omega,\eta\in\bigwedge V$ are elements of pure degree, then $\omega\cdot\eta=(-1)^{|\omega||\eta|}\eta\cdot\omega$ (the multiplication here is the exterior product, I use bars to denote the degree).
Now, in general, if you have two graded $k$-algebras $R=\bigoplus_{i\ge0}R_i$ and $S=\bigoplus_{j\ge0}S_j$, then their tensor product $R\otimes_kS=\bigoplus_{k\ge0}\bigoplus_{i+j=k}R_i\otimes_kS_j$ is naturally a graded $k$-vector space with the indicated grading. In fact, $R\otimes_kS$ can be naturally made into a graded $k$-algebra itself, but here is the surprise. If $r\in R_i,s\in S_j,r^{\prime}\in R_{i^{\prime}}$ and $s^{\prime}\in S_{j^{\prime}}$, we ought to define $(r\otimes s)\cdot(r^{\prime}\otimes s^{\prime})=(-1)^{ji^{\prime}}(r\cdot r^{\prime})\otimes(s\cdot s^{\prime})$. This is in accordance with a general principle known as the "Koszul sign rule", which says that when you exchange two graded symbols of degree $j$ and $i^{\prime}$ ($s$ and $r^{\prime}$ here), the sign $(-1)^{ji^{\prime}}$ ought to be inserted. This makes $R\otimes_kS$ into a graded $k$-algebra, which we call the graded tensor product of $R$ and $S$.
Here's a concrete reason why this is the "right" definition: If $R$ and $S$ are graded commutative, then $R\otimes_kS$ is graded commutative too and it would not be if you omitted the sign (check this). Next, if $T$ is any graded commutative $k$-algebra, the two natural $k$-algebra homomorphisms $R\rightarrow R\otimes_kS,\,r\mapsto r\otimes1$ and $S\rightarrow R\otimes_kS,\,s\mapsto1\otimes s$ induce a bijection between pairs of graded $k$-algebra homomorphisms $R\rightarrow T$ and $S\rightarrow T$ and $k$-algebra homomorphisms $R\otimes_kS\rightarrow T$ in the way one would expect (check this and you'll see the sign is crucial here as well). In other words, the graded tensor product is the coproduct in the category of graded commutative $k$-algebras. This generalizes how the (ungraded) tensor product works to the graded setting.
If we now return to the scenario at hand, this deliberation tells us that $\bigwedge V\otimes_k\bigwedge W$ ought to be understood not as a tensor product of algebras, but as a graded tensor product of graded $k$-algebras. The additional sign in the definition of the multiplication in a graded tensor product is precisely the sign that is missing in the equations you write down, so the natural isomorphism $\bigwedge V\otimes_k\bigwedge W\rightarrow\bigwedge(V\oplus W)$ does in fact become an isomorphism of graded commutative $k$-algebras.
Of course, I could've just directly told you to add a sign to the multiplication in $\bigwedge V\otimes_k\bigwedge W$ that magically makes everything work, but I hope this explanation, even if it may not be entirely transparent, highlights that this is not some ad hoc fix, but fundamentally the right perspective on the issue.
Best Answer
$ \newcommand\Ext{\mathop{\textstyle\bigwedge}} \newcommand\MVects[1]{\mathop{\textstyle\bigwedge^{\mkern-2mu#1}}} \newcommand\C{\mathbb C} \newcommand\R{\mathbb R} $To answer question 2:
We need to be really careful. If we let $\Re$ be the operator which takes an $d$-dimensional complex vector space $W$ (with conjugation) to a $2d$-dimensional real vector space $\Re W$ then the following is false: $$ \Re\MVects kV_\C \color{red} \cong \bigoplus_{p+q=k}\MVects{p,q}V $$ but the following is true $$ \MVects k\Re V_\C \cong \bigoplus_{p+q=k}\MVects{p,q}V. $$ To see that the first line is false, just take $k = 0$: the LHS is $\Re\C$ which has real dimension 2 but the RHS is just $\R$. The second line is actually a consequence of an algebra isomorphism $$ \Ext(U\oplus V) \cong \Ext U\mathbin{\hat\otimes}\Ext V $$ where $U, V$ are vector space over the same field and $\hat\otimes$ is the $\mathbb Z/2\mathbb Z$-graded tensor product. I will not worry about proving this however, and instead focus on the exterior powers.
You seem to understand that we can get a map $\varphi^{p,q} : \MVects{p,q}V \to \MVects{p+q}\Re V_\C$ by just using the exterior product: $$ \varphi^{p,q}(v_1\wedge\dotsb\wedge v_p\otimes w_1\wedge\dotsb\wedge w_q) = v_1\wedge\dotsb\wedge v_P\wedge w_1\dotsb\wedge w_q. $$ (An aside: because we're talking about $\Re V_\C$, the expression $v\wedge(iv) \ne 0$! This is because we are only using an $\R$-linear $\wedge$ over $\Re V_\C = V^{1,0} \oplus V^{0,1}$.) We can package these maps together into a map $\varphi : \bigoplus_{p+q=k}\MVects{p,q}V \to \MVects kV_\C$ where $$ \varphi(\sum_{p+q=k}X_{p,q}) = \sum_{p+q=k}\varphi^{p,q}(X_{p,q}) $$ for any $X_{p,q} \in \MVects{p,q}V$. So what we want to do now is show that this is an isomorphism. This requires showing both injevtivity and surjectivity, or one just one of these as long as we make sure the dimensions match. Let's show surjectivity and then count dimensions.
Now that we have a map $\varphi^{-1} : \MVects k\Re V_\C \to \bigoplus_{p+q=k}\MVects{p,q}V$, because the direct sum comes with a projection $\psi^{a,b} : \bigoplus_{p+q=k}\MVects{p,q}V \to \MVects{a,b}V$ where $a+b=k$, the projection $\pi^{a,b}$ is the composition of these maps: $$ \pi^{a,b} : \MVects k\Re V_\C \overset{\varphi^{-1}}\longrightarrow \bigoplus_{p+q=k}\MVects{p,q}V \overset{\psi^{a,b}}\longrightarrow \MVects{a,b}V. $$