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Let $A$ be a ring, $M$ an $A$-module and $M_1,M_2$ two flat $A$-submodules of $M$. Is $M_1 \cap M_2$ a flat $A$-submodule of $M$?

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No. Pick any flat $A$-module $F$ admitting a non-flat submodule $N\subset F$, and take $M$ to be the pushout of $N\hookrightarrow F$ with itself, i.e. $M := \text{coker}(N\xrightarrow{\Delta} F\oplus F)$, i.e. the "glueing" of two copies of $F$ along $N$. Then $M$ has two copies of the flat module $F$ as submodules, and their intersection is the non-flat module $N$.

To find an example of a non-flat submodule of a flat module, you may restrict to finitely generated modules over a Noetherian ring $A$, where flatness and projectivity are the same, and pick any non-projective submodule of a projective module - for that, see Is every submodule of a projective module projective?

Addendum Concretely, the answer of user26857 gives an example for the above.

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No. Let $R=K[X,Y]$, $I=(X,Y)$, $L=\{(r,-r)\mid r\in I\}$, $M=R^2/L$, $\pi:R^2\to M$, $M_1=\pi(R\times\{0\})$ and $M_2=\pi(\{0\}\times R)$. Then $M_i$ are isomorphic to $R$, while $M_1\cap M_2$ is isomorphic to... (A proof that $I$ is not $R$-flat can be found in this answer.)

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