I thought is worth adding a reference to this:

Miyata, Takehiko, Note on direct summands of modules, J. Math. Kyoto Univ. 7 1967 65–69.

In the paper, the question of the OP is attributed to Matsumura and the solution to Toda. Then a short argument is given for this case.

The author generalizes this result to the following (which is a quote from MathSciNet):

"Let $R$ be a commutative, noetherian ring and let $A$ be an $R$-algebra of finite type. Moreover, let $M$ be a finitely generated $A$-module and let $N$ be a submodule of $M$. Using the usual tools of homological algebra and noetherian ring theory, the author establishes the following pair of results.

Theorem 1: If $M$ is isomorphic to $N\oplus M/N$, then $N$ is a direct summand of $M$.

Theorem 2: If $0\to N\otimes_R T\to M\otimes_R T$ is exact for all $A$-modules $T$ (i.e., $N$ is pure), then $N$ is a direct summand of $M$."

I thought it worth adding a reference to this:

Miyata, Takehiko, Note on direct summands of modules, J. Math. Kyoto Univ. 7 1967 65–69.

In the paper, the question of the OP is attributed to Matsumura and the solution to Toda. Then a short argument is given for this case.

The author generalizes this result to the following (which is a quote from MathSciNet):

"Let $R$ be a commutative, noetherian ring and let $A$ be an $R$-algebra of finite type. Moreover, let $M$ be a finitely generated $A$-module and let $N$ be a submodule of $M$. Using the usual tools of homological algebra and noetherian ring theory, the author establishes the following pair of results.

Theorem 1: If $M$ is isomorphic to $N\oplus M/N$, then $N$ is a direct summand of $M$.

Theorem 2: If $0\to N\otimes_R T\to M\otimes_R T$ is exact for all $A$-modules $T$ (i.e., $N$ is pure), then $N$ is a direct summand of $M$."