Why is the cofree coalgebra defined in this way? According to wikipedia, the cofree coalgebra (not requiring it to be a bialgebra) is defined with the coproduct (over v) defined as
$$\Delta: V\to V\boxtimes V$$
$$\Delta: v \mapsto v\boxtimes 1 + 1\boxtimes v$$
$$ 1 \mapsto 1 \boxtimes 1$$
A more natural definition (for me) would be 
$$\Delta: v \mapsto v\boxtimes v$$
Why not define it this way? Is it because to make it a graded morphism? That's the only reason I can think of, given that you can define a linear morphism over a basis pretty much any way you want. Is respecting the grading such a big deal? 
 A: It passed a long time since this question was asked and the question itself contains an easy-to-find mistake as it has been highlighted in the comments. Nevertheless, I think that it deserves an answer anyway.
Let $\Bbbk$ be a field, just to stay close to earth. The main point is that the term cofree coalgebra has a very precise meaning: it denotes the right adjoint (if it exists!) to the underlying functor $$U:\mathrm{Coalg}_{\Bbbk}\to\mathrm{Vec}_{\Bbbk}.$$
Usually, when you speak about free objects, you are looking for a left adjoint (as it happens e.g. with the free associative algebra $T_{\Bbbk}(V)$ over a vector space $V$: the functor $T_{\Bbbk}:\mathrm{Vec}_{\Bbbk}\to\mathrm{Alg}_{\Bbbk}$ is the left adjoint to the forgetful functor). However, as the name says, coalgebras are dual constructions to algebras, so you may expect everything to behave dually.
Now, for every vector space $V$, the cofree coalgebra on $V$ is a coalgebra $C(V)$ together with a canonical $\Bbbk$-linear morphism $\pi:C(V)\to V$ such that for every $\Bbbk$-coalgebra $D$ and every $\Bbbk$-linear map $f:U(D)\to V$ there exists a unique coalgebra morphism $F:D\to C(V)$ that satisfies $\pi\circ F=f$. This free coalgebra always exists, but its explicit construction is very tricky in my opinion (you may have a look at Sweedler's Hopf algebras §6.4, Abe's Hopf algebras §4.2 or here). In a nutshell, the idea is the following.
First of all, for every vector space $V$, you consider the finite dual $T_\Bbbk(V^*)^\circ$ of the free associative algebra $T_\Bbbk(V^*)$ over the dual vector space $V^*$. It is a coalgebra by construction and it comes endowed with a canonical morphism $\pi:T_\Bbbk(V^*)^\circ\to V^{**}$ which makes of it the cofree coalgebra on $V^{**}$.
As a second step, observe that $V$ can be identified with a subspace of $V^{**}$ via the evaluation map $\mathrm{ev}:V\to V^{**};v\mapsto \mathrm{ev}_v$, where $\mathrm{ev}_v(f)=f(v)$. Then consider the biggest subcoalgebra $C\subseteq T_\Bbbk(V^*)^\circ$ such that $\pi(C)\subseteq V$: this can be taken as the sum of all subcoalgebras $D\subseteq T_\Bbbk(V^*)^\circ$ such that $\pi(D)\subseteq V$. The restriction of $\pi$ to $C$ makes of it the cofree coalgebra $C(V)$ on $V$.
Since it exists, it is unique up to isomorphism (because it satisfies a universal property) and this is the reason why you are not exactly free to choose the comultiplication.
Even if you fix a basis of $V$, define $\Delta$ as the group-like comultiplication on the elements of this basis and then extend this $\Delta$ by linearity to a suitable comultiplication, what may happen is that the coalgebra you find at the end of the procedure is not the cofree coalgebra, but some other coalgebra you may associate to $V$.
For example, notice that if $V=\mathrm{Span}_\Bbbk\{e_1,\ldots,e_n\}$ and you define $\Delta$ and $\varepsilon$ by letting $\Delta(e_i)=e_i\otimes e_i$ and $\varepsilon(e_i)=1$ extended by linearity (as we were saying above) then $(V,\Delta,\varepsilon)$ is in fact a coalgebra associated to $V$, but it is not the cofree coalgebra.
