Let's work on Baire space $\omega^\omega$. $\mathrm{cof}(\mathcal{C},\mathcal{M})$ is the least size of a collection of meager sets so that any countable closed set is a subset of the member of the collection. $\mathrm{cof}(\mathcal{C}',\mathcal{M})$ is the same, but where $\mathcal{C}'$ is all possible countable sets. $\mathrm{cov}(\mathcal{M})$ is the least size of a collection of meager sets whose union equals $\omega^\omega$.
I claim that all three numbers are equal. In other words, in terms of cardinality it doesn't matter whether we ask for a collection of meager sets cofinal over the countable sets, the closed countable sets, or even just the one element sets.
$\mathrm{cof}(\mathcal{C},\mathcal{M})\leq \mathrm{cof}(\mathcal{C}',\mathcal{M})$ is straight from the definition.
$\mathrm{cov}(\mathcal{M})\leq\mathrm{cof}(\mathcal{C},\mathcal{M})$ is almost as straightforward; since sets of size 1 are closed any collection of meager sets cofinal in the countable closed sets covers the whole space.
$\mathrm{cof}(\mathcal{C}',\mathcal{M})\leq \mathrm{cov}(\mathcal{M})$: We use a characterization of $\mathrm{cov}(\mathcal{M})$ due to Bartoszynski (this is Theorem 2.4.1 in his book with Judah): it is the least size of a family $\mathcal{F}\subseteq\omega^\omega$ so that any $g\in\omega^\omega$ is eventually different from some $f\in\mathcal{F}$, meaning $(\forall^\infty n)f(n)\not=g(n)$. So fix such a family. Also fix a partition of $\omega$ into countably many infinite pieces $\{A_k:k\in\omega\}$. For each $f\in\mathcal{F}$ let $M_f$ be the set of all $x\in\omega^\omega$ for which there exists a $k$ so that $(\forall^\infty n\in A_k)f(n)\not=x(n)$. Then $M_f$ is meager. We claim the collection of $M_f$ is cofinal over the countable sets.
Fix $C=\{h_k:k\in\omega\}$ a countable set. Let $g_C\in\omega^\omega$ be given by $g_C(n)=h_k(n)$ if $n\in A_k$. There is some $f\in\mathcal{F}$ so that $g_C$ is eventually different from $f$. And this implies that $C\subseteq M_f$. QED.
