Let $(A_i)_{i \in I}$ be a family of objects in your category. Now, the cartesian product $\prod_i A_i$ has a $G$-set structure via $g(x_i)_{i \in I} = (gx_i)_{i \in I}$ which makes the projections $\pi_j : \prod_i A_i \to A_j$ to be $G$-equivariant. Now, let's see that $\prod_i A_i$ with these arrows is the product of $(A_i)_{i \in I}$ by verifying the universal property of the product. If you have arrows $h_j : X \to A_j$, then
$$
h(x) := (h_i(x))_{i \in I}
$$
is an arrow $h: X \to \prod_j A_j$ since for each $g \in G$,
$$
gh(x) = g(h_i(x))_{i \in I} = (gh_i(x))_{i \in I} = (h_i(gx))_{i \in I} = h(gx).
$$
It is also clear from the definition of $h$ that $\pi_j h = h_j$ for all $j$, and this is the only arrow that satisfies this propery: if $h' : X \to \prod_iA_i$ is such that $\pi_jh' = h_j$, then
$$
h'(x)_j = \pi_jh'(x) = h_j(x) = h(x)_j \quad (\forall j \in I)
$$
and so $h'(x) = h(x)$ for all $x \in X$. Thus our candidate verifies the universal property of the product, which makes it the product of the $A_i$ (up to isomorphism, of course). Here we take advantage of the fact that our objects and arrows have an underlying structure as sets and functions.
As for the final object: again, our objects have an underlying set structure, so a natural candidate is a one element set $\{*\}$. Note that any other one element $G$-set is isomorphic to $\{*\}$ but this is not in contradiction with $\{*\}$ being final. In effect, if $X$ is any $G$-set, the constant map $c:X \to *$ is $G$-equivariant, and it is the only possible arrow $X \to *$ since it is the only existing function from $X$ to $\{*\}$ to begin with.
First of all, that wikipedia definition is not great. You should however go back to it once your wrap your head around all this stuff.
When one first encounters the definition of a universal construction, it's a bit weird. The true way to understand this concept is with examples: just keep looking at examples until it makes sense.
Let's start with a simple example. Let $X, Y$ be sets. Then (one possible way) I can define the disjoint union is
$$
X \amalg Y = \{(x, 0), (y, 1) \mid x \in X, y \in Y \}
$$ and we can define injection morphisms
$$i_X: X \to X \amalg Y \qquad i_X(x) = (x, 0)\\
i_Y: Y \to X \amalg Y \qquad i_Y(y) = (y, 1)
$$
Now suppose I have a function $f: X \to Z$ and $g: Y \to Z$. Then I can construct a unique (this is really key here) map $h: X \amalg Y \to Z$ where
$$
h(x, 0) = f(x) \text{ and } h(y, 1) = g(y)
$$
Why is it unique? Because the definition depends directly on $f$ and $g$. So what this really gives me is this diagram
Given the arrows $f:X \to Z, g: Y \to Z$, we can get a unique arrow $h: X \amalg Y \to Z$. The forced existence of $h$ is indicated by the dashed arrow.
This is an example of a universal construction. Actually, this is an example of a colimit, which I will explain.
This happens in so many places in math that there's a name for it, and thats what this universal arrow stuff is about.
So suppose $\mathcal{C}$ and $\mathcal{D}$ are categories with a functor $F: \mathcal{C} \to \mathcal{D}$. A universal morphism from an object $D$ to the functor $F$ is a pair $(C, u: D \to F(C))$ such that, for any $f: D \to F(C')$, the following diagram holds.
So if you have a morphism $f: D \to F(C')$, you automatically get a morphism $h: C \to C'$.
A colimit is an example of this construction. To understand this, you first need to understand the diagonal functor
$$
\Delta: \mathcal{C} \to \mathcal{C}^J.
$$
Here, $\mathcal{C}^J$ is the functor category of functors $F: J \to C$, with morphisms as natural transformations. This functor $\Delta$ takes each object $C$ to the functor $F_C: J \to C$, where for each $j \in J$
$$
F(j) = C.
$$
So it sends it to a constant valued functor.
Now when we speak of a "colimit" in a category $\mathcal{C}$
it's with respect to some functor $F: J \to C$. This is an element of the functor category $\mathcal{C}^J$. So, we define a colimit to be a universal morphism from $F$ to $\Delta$. That is,
$$
(\text{Colim }F, u: F \to \Delta(\text{Colim} F).
$$ Note that $u$ is natural transformation; as I said earlier $\Delta$ sends objects to functors. So, you get the diagram
However, this isn't very intuitive. A better way to look at this is to realize that if you have a natural transformation $u: F \to \Delta(\text{Colim } F)$,
then you have a family of morphisms. How so? For each object $i \in J$, our natural transformation should give us a morphism
$$
u_i: F(i) \to \Delta(\text{Colim } F)(i).
$$
But $\Delta(\text{Colim } F)$ is a constant valued functor. So this really becomes a family of morphisms
$$
u_i: F(i) \to \text{Colim }F.
$$
This is usually how people define colimits, but since you asked how they're related to universal constructions, this is how. Anyways, one way to imagine the above diagram is to picture the one below. This isn't exactly precise, but it's a good way to imagine the colimit.
Compare this diagram with the one in the beginning with the coproduct.
If you understand that, then you can understand the concept of limits because limits are the same exact story, just with the arrows reversed.
Best Answer
This is indeed a universal property, and we can indeed find a category where $(V, \phi)$ is initial.
The category that you're looking for is a certain kind of comma category.
Here's a direct definition.
Objects are pairs $(U, g)$, where $U$ is a vector space and $g$ is an ordinary function $g : I \to U$.
An arrow from $(U, g)$ to $(W, h)$ is a linear map $T : U \to W$ wuch that $h = T \circ g$.
Composition is just ordinary composition of functions. It's easy to verify the axioms of a category from here.
Then we see that the claim "$V$ is the free vector space on $I$" is simply stating that $(V, \phi)$ is the initial object.
This is related to notions of the representability of a functor and of adjoints.
Edit: if you insist that a universal property be stated in terms of a terminal object, just take the opposite category of the one I have described, which makes $(V, \phi)$ into a terminal object.