Let $f: S^k \longrightarrow S^r$ be a smooth map. Then its suspension $Sf: S^{k+1} \longrightarrow S^{r+1}$ is smooth away from the basepoints (i.e. the north of south poles of $S^{k+1}$ and $S^{r+1}$). Let $x \in S^r$ be a regular value of $f$ different than the basepoint of $S^r$, and let $A = f^{-1}(x)$ be the framed submanifold of $S^k$ associated to $f$. We clearly have that $A = (Sf)^{-1}(x) \subset S^k \subset S^{k+1},$ where we consider $S^k$ as the equator of $S^{k+1}$ and view $x \in S^r$ as lying in the equator of $S^{r+1}$. Now \begin{align*} \nu(A \hookrightarrow S^{k+1}) & = \nu(A \hookrightarrow S^k) \oplus \nu(S^k \hookrightarrow S^{k+1}) \\ & = \nu(A \hookrightarrow S^k) \oplus \varepsilon_A^1, \end{align*} where $\varepsilon_A^1$ is the trivial line bundle over $A$, and similarly \begin{align*} \nu(\{x\} \hookrightarrow S^{r+1}) & = \nu(\{x\} \hookrightarrow S^r) \oplus \nu(S^r \hookrightarrow S^{r+1}) \\ & = \nu(\{x\} \hookrightarrow S^r) \oplus \varepsilon_{\{x\}}^1. \end{align*} Since near the equator we have $Sf = f \times \mathrm{Id}$, $Sf$ preserves the canonical framing of $S^r$ in $S^{r+1}$, and therefore we see that we get a well-defined framing of $A$ in $S^{k+1}$.
From the above, we see that when $M \cong S^k$, the Pontrjagin-Thom construction tells us that suspensions of homotopy classes of maps $f: S^k \longrightarrow S^r$ correspond to framed submanifolds of $S^k$ embedded in the equator of $S^{k+1}$, with its framing Whitney summed with the canonical framing of $S^k$ in $S^{k+1}$.