PS: The answer below has gaps, and it is likely incorrect.
Yes, $C^{\infty}(\mathbb{R}^{m+n})$ is a flat $C^{\infty}(\mathbb{R}^{m})$ module. Or, following @Pietro 's comment, with more generality, if $X,Y$ are smooth manifolds, then $C^\infty(X\times Y)$ is a $C^\infty(X)$ flat module. To see this, it is enough to consider $C^\infty(X)$ linear, injective maps $\varphi: K \rightarrow L$, where $K = <r_1,\dots,r_d>_{C^\infty(X)}$ is a finitely generated $C^\infty(X)$ ideal, and $L=C^\infty(X)$ is the ring of scalars.
Notice that in that case, considering that $C^\infty(X)$ is a sub-ring of $C^\infty(X\times Y)$, the tensor operation $.\otimes_{C^\infty(X)} C^\infty(X\times Y) $ applied to an ideal such as $K$ is just the extension of scalars, i.e., in some natural way, $$ K\otimes_{C^\infty(X)} C^\infty(X\times Y) = <r_1,\dots,r_d>_{C^\infty(X\times Y)}, $$ and $$L\otimes_{C^\infty(X)} C^\infty(X\times Y) = C^\infty(X\times Y).$$ Moreover, with these representations, $\bar{\varphi} = \varphi \otimes Id$, the extension of $\varphi$ from $K$ to $K\otimes C^\infty(X\times Y)$, satisfies: $$ \bar{\varphi}( \sum_{i=1}^d r_i\ m_i) = \sum_{i=1}^d \varphi(r_i)\ m_i $$ for any $m_1,\dots, m_d \in C^\infty(X\times Y)$.
We want to verify that, under the assumption the $\varphi$ is injective, $\bar{\varphi}$ is injective. Now, if for some linear combination $\sum_i r_i \ m_i$ we have $\bar{\varphi}(\sum_i r_im_i) = 0$, then $\sum_i \varphi(r_i)\ m_i = 0$ in $C^\infty(X\times Y)$, and so, for all $y \in Y$, $$ \sum_i \varphi(r_i)\ m_i(.,y) = 0 $$ in $C^\infty(X)$, but $\varphi$ being $C^\infty(X)$ linear, we get: $$ \varphi( \sum_i r_i\ m_i(.,y) ) = 0. $$ By injectivity of $\varphi$, $ \sum_i r_i\ m_i(.,y) = 0$
(With @Pietro notation, this is the condition of an element of $M^Y$ having all 0 coordinates applied to the case $M = $ a finitely generated ideal of $C^\infty(X)$)
Since this holds for all $y$, then $\sum_i r_i\ m_i = 0$, which shows injectivity of $\varphi$.