Let $R$ be a local ring with maximal ideal $I$. $M$ is a finitely generated module over $R$ generated by $a_1, \ldots, a_n$ and the generators are chosen such that their quotients in $M/IM$ form a basis. Then there is a surjective homomorphism $f: R^{n} \to M$. Suppose that $R^{m} = M \oplus \ker f$. How to show that $\ker f = I\ker f$?
Suppose $(p_1, \ldots, p_n) \in \ker f$, then $p_1a_1+\ldots+p_na_n = 0$. It then implies that $p_1a_1+\cdots+p_na_n = 0$ in $M/IM$. So $p_i \in I, \forall i$. But this is not enough to conclude that $\ker f = I\ker f$.
Put $N = \mathrm{ker}(f)$. Since $M$ is projective, the short exact sequence
$$0 \to N \to R^{n} \xrightarrow{~f~} M \to 0$$
is split, and so there is an isomorphism $\alpha \colon R^{n} \to M \oplus N$ witnessing this splitting. (This is all we need below, but it is worth noting that if $R^{m} \cong M \oplus N$, then $R^{m} \cong R^{n}$ via $\alpha$, whence $m = n$, since all commutative rings satisfy the invariant basis property.)
Now, let $k = R/I$ be the residue field of $R$. Then $\alpha \otimes \mathrm{Id}_{k} \colon R^{n} \otimes_{R} k \to (M \oplus N) \otimes_{R} k$ is an isomorphism of $k$-vector spaces. Since $R^{n} \otimes_{R} k \cong k^{n}$ as $k$-modules, it follows that $\dim_{k}((M \oplus N) \otimes_{R} k) = n$. But $(M \oplus N) \otimes_{R} k \cong M/IM \oplus N/IN$ as $k$-modules, and since $\dim_{k}(M/IM) = n$, we have $\dim_{k}(N/IN) = 0$. Hence, $N/IN = 0$, whence $N = IN$, as desired.