I'm surprised nobody has answered yet so let me state (what seems to me to be) the obvious: (2) is the only reasonable definition of sub-coalgebra which makes sense in general. Said differently you want $\Delta:D\rightarrow C\otimes C$ to factor through $D\otimes D \rightarrow C\otimes C$. The definition in your notes is either an oversight or an abuse of language, I'd say.

As მამუკა ჯიბლაძე suggests, coalgebras in $R$-mod are the same as algebras in $R-mod^{op}$, and you don't require the latter map to be injective in the same way that for a quotient algebra $A\rightarrow B$ you don't usually require $A\otimes A\rightarrow B\otimes B$ to be surjective as well (why would you ?). Of course for algebras over a commutative ring this is automatically true, but it's not true for symmetric monoidal categories whose tensor product is not right exact, which is exactly what happens for $R-mod^{op}$.

The downside is that the coalgebra structure on a sub-coalgebra might not be uniquely determined, i.e. it is possible to have two different coalgebra structures on $D$ for which the inclusion $D\rightarrow C$ is a coalgebra map. In other words, the factorisation of the map $D\rightarrow C\otimes C$ might not be unique. Somewhat related, if $C$ is not flat then its category of comodules is not abelian.

See e.g. Wischnewsky, On linear representations of affine groups. I. for a reference where this is carefully phrased.

I'm surprised nobody has answered yet so let me state (what seems to me to be) the obvious: (2) is the only reasonable definition of sub-coalgebra which makes sense in general. Said differently you want $\Delta:D\rightarrow C\otimes C$ to factor through $D\otimes D \rightarrow C\otimes C$. The definition in your notes is either an oversight or an abuse of language, I'd say.

As მამუკა ჯიბლაძე suggests, coalgebras in $R$-mod are the same as algebras in $R-mod^{op}$, and you don't require the latter map to be injective in the same way that for a quotient algebra $A\rightarrow B$ you don't usually require $A\otimes A\rightarrow B\otimes B$ to be surjective as well (why would you ?). Of course for algebras over a commutative ring this is automatically true, but it's not true for symmetric monoidal categories whose tensor product is not right exact, which is exactly what happens for $R-mod^{op}$.

The downside is that the coalgebra structure on a sub-coalgebra might not be uniquely determined, i.e. it is possible to have two different coalgebra structures on $D$ for which the inclusion $D\rightarrow C$ is a coalgebra map. In other words, the factorisation of the map $D\rightarrow C\otimes C$ might not be unique. Somewhat related, if $C$ is not flat then its category of comodules might not be abelian.

See e.g. Wischnewsky, On linear representations of affine groups. I. for a reference where this is carefully phrased.