Question: Let $f,g: \mathbb{R}^n \to \mathbb{R}^m$, $f$ and $g$ are transversals if $f(x) = g(x)$ implies $\Im(Df_x) + \Im(Dg_x) = \mathbb{R}^m$.
Let $R: \mathbb{R}^m \to \mathbb{R}^m $ and $S: \mathbb{R}^n \to \mathbb{R}^n$ diffeomorphisms, suppose $R \circ f \circ S^{-1}(x)= R \circ g \circ S^{-1}(x)$ and show that $\Im(D (R \circ f \circ S^{-1})_x) + \Im( D(R \circ g \circ S^{-1})_x) = \mathbb{R}^m$.
Let $A$ and $B$ two linear transformations from $\mathbb{R}^n\rightarrow \mathbb{R}^m$ such that $Im(A)+Im(B)=\mathbb{R}^m$. Then if $S : \mathbb{R}^n\rightarrow \mathbb{R}^n$ is an invertible linear transformation it is basic linear algebra to check that $Im(A \circ S)+Im(B \circ S)=\mathbb{R}^m$. In the same spirit, if $R:\mathbb{R}^m\rightarrow \mathbb{R}^m$ is an invertible linear transformation we have that $Im(R \circ A)+Im(R \circ B)=\mathbb{R}^m$. Combining these two results, we have that $Im(R \circ A \circ S^{-1})+Im(R \circ B \circ S^{-1})=\mathbb{R}^m$
Now, going back to your problem : as $R$ and $S$ are diffeomorphisms, $DR$ and $DS^{-1}$ are everywhere invertibles. Hence you can apply what we just saw : we have that $f(S^{-1}(x))=g(S^{-1}(x))$ hence because $f$ and $g$ are transversals we have that $$Im(Df_{S^{-1}(x)})+Im(Dg_{S^{-1}(x)})=\mathbb{R}^m$$
Hence $$Im(DR_{f(S^{-1}(x))}\circ Df_{S^{-1}(x)}\circ DS^{-1}_x)+Im(DR_{g(S^{-1}(x))}\circ Dg_{S^{-1}(x)}\circ DS^{-1}_x)=\mathbb{R}^m$$
that is $$Im(D(R\circ f\circ S^{-1})_x)+Im(D(R\circ g\circ S^{-1})_x)=\mathbb{R}^m$$