What does it mean for a coalgebra to be cogenerated by a subspace? The usual definition of an algebra being generated by a subspace is as follows:

Let $A$ be an algebra, $X \subset A$ a subspace, $\mathrm{Alg}(X)$ the free algebra generated by $X$. Then $A$ is generated by $X$ iff the morphism $\mathrm{Alg}(X) \to A$ induced by the inclusion $X \hookrightarrow A$ is a surjection.

The interpretation here is clear: every element of $A$ can be written as a finite linear combination of "words" of elements of $X$.
Now the dual statement is (see eg. Operads in algebra, topology and physics by Markl et al.):

Let $C$ be a coalgebra, $Y$ a quotient of $C$, $\mathrm{Alg^c}(Y)$ the cofree coalgebra cogenerated by $Y$. Then $C$ is cogenerated by $Y$ if the morphism $C \to \mathrm{Alg^c}(Y)$, coinduced by $C \twoheadrightarrow Y$, is injective.

I don't know how to interpret that. If we restrict ourselves to conilpotent coalgebras, we can take the cofree conilpotent coalgebra generated by $Y$ to be $T^c(Y)$ the usual tensor coalgebra; $T^c(Y) \to Y$ is the projection on the $Y$ factor, and we have an explicit description of the morphism $\tilde p : C \to T^c(Y)$, with the projection of the $Y^{\otimes n}$ factor is $\sum p(x_{(1)}) \otimes \dots \otimes p(x_{(n)})$. Somehow if $Y$ cogenerates $C$, then this is injective. Which would roughly mean that for any $x \in C$, if you iterate the coproduct enough time, the factors are nonzero through the projection on $Y$. What does that mean? Is there a way to interpret that?
 A: To understand cogeneration for coalgebras it helps to really understand generation for algebras, so lets do that first. We are going to work in the category of vector spaces, which we write as ${\bf Vec}$. 
An algebra will be a vector space $V$ together with maps $V \otimes V \to V $ and $ k \to V $ which satisfy the associative and unital laws. Write ${\bf Alg}$ for the category of algebras. We have a forgetful functor $ U : {\bf Alg} \to {\bf Vec}$. The functor $U$ preserves limits. Modulo set theoretic issues, continuious functors always have left adjoints. Therefore the functor $U$ has a left adjoint $ F : {\bf Vec} \to {\bf Alg}$ which we call the free functor. The counit of this adjunction is surjective, so every algebra $A$ admits a surjection $F(V) \to A $ for some vector space $V$. Luckily for humanity, the algebra $F(V)$ is quite familiar. It is the tensor algebra on $V$, which is like a polynomial algebra except we no longer have $X_i X_j = X_j X_i $. If we have a surjection $\pi : F(V) \to A$ we say that $ \pi(V) $ generates $A$. 
Now lets do coalgebras. A coalgebra is a vector space $V$ together with maps $V \to V \otimes V$ and $ V \to k$ which satisfy the coassociative and counital laws. Write ${\bf coAlg}$ for the category of coalgebras. Again we have the forgetul functor $ U : {\bf coAlg} \to {\bf Vec}$. But, now $U$ preserves all colimits. Therefore $U$ has a right adjoint $C : {\bf Vec} \to {\bf coAlg}$ which we call the cofree functor. The left adjoint $U$ is faithful, which implies that the unit of the adjunction is injective. Therefore, if $A$ is a coalgebra, we have an injection $A \hookrightarrow CU(A)$. In particular, every coalgebra embeds into a cofree coalgeba. Unfortunately for humanity, cofree coalgebras are not easy to work with. To get a feel for what I mean, you should check out the cofree coalgebra nlab page. If you have an injection $A \hookrightarrow C(V) $, then you say that $A$ is cogenerated by $V$, but since cofree coalgebras not easy to work with this is not such a useful notion.
All is not lost. A lot of the time when coalgebras come up in everyday mathematics, they have some extra structure. For myself, this extra structure is usually a grading. Just as graded algebras are easier to work with than than arbitrary algebras, graded coalgebras are easier to work with than arbitrary coalgebras, so lets talk about them. By a graded coalgebra, we mean a coalgebra $A$ equipped with a grading $A = \oplus_{d \geq 0} A_d $ such that


*

*The comultiplication takes $A_d$ into $\oplus_{p+q=d} A_p \otimes A_q$

*$A_0 = k$ and the counit is given by quotienting with $A_+$


Here is an example of a graded coalgebra. Let $V$ be a vector space. Define $$T(V) = \oplus_{d \geq 0} V^{\otimes d}.$$
This has a comultiplication defined by
$$ v_1\cdots v_n \mapsto \sum_{i=0}^n v_1 \cdots v_i \otimes v_{i+1} \cdots v_n. $$ 
Let $A$ be a graded coalgebra and write $ \Delta$ for the comultiplication. We write
$$ \overline{\Delta}(x) = \Delta(x) - x \otimes 1 - 1 \otimes x $$ 
This is called the reduced comultiplication and is coassociative (which
you can prove by induction). Notice that the reduced comultiplication
of any element in $ A_1 $ is $ 0 $. More generally, we have that
$ \overline{\Delta}^n(a) = 0 $ for $ a \in A_n $. If high enough reduced copowers of every element are zero, then we call $A$ conilpotent. 
It often happens that a coalgebra appearing in nature is conilpotent. If we consider the forgetful functor $ U : {\bf coNilcoAlg} \to {\bf
  Vec}$ then it has a right adjoint which is given by $ V \mapsto
T(V) $. Therefore, for a conilpotent coalgebra, we can say that it is
cogenerated by $ V $ if it embeds in $ T(V) $, which is
much better than what we can say for arbitrary coalgebras.
A: I can't comment under DBr's answer for being a noob. But at least in the conilpotent case (maybe also pointed coalgebra case), he has answered your question: every element of $C$ is a sum of words in $Y$ freely joined by $\otimes$.
Just think in terms of combinatorics: comultiplication codes the deconcatenation of words ($abcd \to a | bcd + ab | cd + abc | d$ using reduced $\overline{\Delta}$) which is dual to concatenation of words coded by multiplication ($a * bcd \to abcd$ etc). You can think $C$ is formed by words in $Y$ but you are looking at deconcatenation process in the algebraic structure.
Edit: Not every words will be in C so I deleted "freely generated".
