This is an elaboration of Ralph's answer to the question linked by the OP. I use the notation from there. Moreover, I assume the axiom of countable choice, i.e. countable unions of countable sets are countable.

Let $P$ be a non-countably generated projective $R$-module with projective base $(x_i, f_i)_{i\in I}$. A projective base has the two properties:

For all $x \in P$:
$$\text{supp}(x) := \{i \in I \mid f_i(x) \neq 0\} \text{ is finite}\tag{1}$$
$$x = \sum_{i \in\text{supp}(x)}f_i(x)x_i \tag{2}$$

Let $M$ be a submodule of $P$ with countably (finite or infinite) many generators $y_k$. Define inductively
$$I_0 := \bigcup_k \text{supp}(y_k)$$
$$I_{n+1} := \bigcup_{i \in I_n} \text{supp}(x_i)$$
and let $J = \bigcup_{n \ge 0}I_n$. By the axiom of countable choice, the sets $I_n, J$ are countable. Let $Q$ be the submodule of $P$ generated by (the countable many) $x_i,\,i \in J$.

By $(2)$, $y_k$ is an $R$-linear combination of $x_i\,(i \in \text{supp}(y_k)\subseteq J)$ and hence $M \subseteq Q$.

The essential step is to show that $Q$ is a direct summand of $P$. This is done by showing that the following $R$-linear map is a projection (i.e. fixes $Q$ pointwise):

$$\kappa: P \to Q,\,\,\,x \mapsto \sum_{i \in \text{supp}(x)\cap J}f_i(x)x_i$$

Let $x_j$ be a generator of $Q$. That means $j \in J$, i.e. there is $n \ge 0$ such that $j \in I_n$. Hence $\text{supp}(x_j)\subseteq I_{n+1}\subseteq J$ and

$$\kappa(x_j) \overset{\text{def}}{=}\sum_{i \in \text{supp}(x_j)\cap J}f_i(x_j)x_i = \sum_{i \in \text{supp}(x_j)}f_i(x_j)x_i\overset{(2)}{=}x_j.$$
So $\kappa$ is the identity on $Q$ and the inclusion $Q \hookrightarrow P$ is a splitting of $\kappa$.

Finally, $Q$ is a proper submodule of $P$, since $Q$ is countably generated and $P$ is not countably generated.$\blacksquare$