If I understand the question correctly, the only thing you're having issues with is why your underlined term $e_3^*(e_{\sigma(3)})$ is zero if $\sigma(3) = 1,2$ and is $1$ if $\sigma(3) = 3$. But if you go back to the top of your post, this is precisely the definition of the dual basis vector $e_3^*$. Things don't always work out so nicely, of course. You are taking a wedge product of some very simple functions, and applying them to particularly easy vectors.
As for your first question (which I guess your instructor helped you with), you should see that this is just a consequence of the associativity of your wedge product (what you wrote in your first gray box). What you were confused about is why it says $\operatorname{Alt}[(e_1^* \wedge e_2^*) \otimes e_3^*]$ rather than $\operatorname{Alt}(e_1^* \otimes e_2^* \otimes e_3^*)$. But the first expression here is precisely $(e_1^* \wedge e_2^*) \wedge e_3^*$, so these things are equal (I assume your definition of the wedge product is $\alpha \wedge \beta = \operatorname{Alt}(\alpha \otimes \beta)$).
I hope this was helpful for you!
The idea is that the wedge product gives you a canonical application $\mathbb{R}^3\times \mathbb{R}^3\to \bigwedge^2(\mathbb{R}^3)$, and that there is then a more-or-less canonical isomorphism $\bigwedge^2(\mathbb{R}^3)\to \mathbb{R}^3$, and composing the two yields a map $\mathbb{R}^3\times \mathbb{R}^3\to \mathbb{R}^3$ which is precisely the cross product.
I think one gets a clearer idea of what's happening by looking at a more general case. Let's say that $E$ is a $3$-dimensional vector space (let's say over the real numbers, but it doesn't matter). Then there is a canonical map $E\times E\to \bigwedge^2(E)$, namely the wedge product. But there is no canonical equivalent of the cross product $E\times E\to E$. That is because although $E$ and $\bigwedge^2(E)$ both have dimension $3$, and are therefore isomorphic, they are not canonically isomorphic.
So what do we need to add to get a natural map $E\times E\to E$, or equivalently a natural isomorphism (an identification, as your question puts it) $\bigwedge^2(E)\simeq E$? What happens is that although there is no natural identification between $E$ and $\bigwedge^2(E)$, there is a natural duality. Namely, we have the wedge product $E\times \bigwedge^2(E)\to \bigwedge^3(E)$, where $\bigwedge^3(E)$ has dimension $1$. If we choose a non-zero element $u\in \bigwedge^3(E)$, we get an identification $\bigwedge^3(E)\simeq \mathbb{R}$, and therefore a (non-degenerate) pairing $E\times \bigwedge^2(E)\to \mathbb{R}$ which defines an isomorphism $\bigwedge^2(E)\simeq E^*$ (where $E^*$ is the dual of $E$). Using this, we get close to our cross product, since the wedge product $E\times E\to \bigwedge^2(E)$ now becomes $E\times E\to E^*$.
To finish, we now need an identification $E\simeq E^*$, which is equivalent to a non-degenerate bilinear form $E\times E\to \mathbb{R}$. You may think of a scalar product (there are other possibilities, but let's say scalar products are more palatable to beginners).
In the end, to transform our initial wedge product $E\times E\to \bigwedge^2(E)$ into a "cross product" $E\times E\to E$, we needed a scalar product on $E$, and a choice of a "volume form" $u\in \bigwedge^3(E)$. In particular, if $E$ is a euclidean space (that is, it has a choice of scalar product), then there are two natural choices for $u$, corresponding to "volume $1$" forms, the two choices corresponding to the two possible orientations of $E$. That means that there is a canonical cross product on any oriented eucliean space of dimension $3$, which is the familiar statement (explicitly, one defines $u\times v$ by the formula
$$ \langle u\times v, w\rangle = \det(u,v,w) $$
for all $w\in E$, where $\det$ is the determinant corresponding to any direct orthonormal basis).
Of course, $\mathbb{R}^3$ is naturally an oriented euclidean space, so we can skip this whole theory and just use the classical formulas. But then we miss what's happening behind the curtain, and it looks like we are just doing random identification, as your question suggests.
Best Answer
One starting comment: instead of $\wedge^3 V$ you want $\wedge^3 \mathbb{R}^3$, since exterior products apply to vector spaces, not sets of basis vectors for them.
The basis for $\wedge^3 \mathbb{R}^3$ is indexed by subsets of $\{1,2,3\}$ of cardinality 3, so there's one basis vector: what you have called $e_{123}$. Thus, it's a one-dimensional vector space. More generally, the dimension of $\wedge^d \mathbb{R}^n$ is the binomial coefficient $\binom{n}{d}$.
The exterior algebra does indeed have the 8 basis vectors you listed above, although $e_\varnothing = 1 \in \mathbb{R}$; this is the "empty" exterior power, so you can think of this as analogous to $c^0 = 1$ for $c \in \mathbb{R}$ (take $c \ne 0$ if you want $0^0$ to be something besides 1 of course!). The exterior algebra contains all of the exterior powers $\wedge^d \mathbb{R}^3$ for $d \ge 0$. For $d > 3$ this is zero, by the binomial dimension calculation earlier.