The Whitehead map is the attaching $(m+n-1)$-cell map for the CW-complex $S^m \times S^n$ (i.e. the map $w: S^{m+n-1} \to S^m \vee S^n $ which can be constructed in the following way: $$S^{m+n-1}=\partial D^{m+n} \approx \partial\left(D^{m} \times D^{n}\right)=\left(D^{m} \times \partial D^{n}\right) \cup\left(\partial D^{m} \times D^{n}\right)=\left(D^{m} \times\right. \left.S^{n-1}\right) \cup\left(S^{m-1} \times D^{n}\right)$$ and $$w:\begin{array}{l} D^{m} \times S^{n-1} \rightarrow D^{m} \rightarrow D^{m} / S^{n-1}=S^{m} \subset S^{n} \vee S^{n} \\ S^{m-1} \times D^{n} \rightarrow D^{n} \rightarrow D^{n} / S^{n-1}=S^{n} \subset S^{n} \vee S^{n} \end{array}).$$
I've found in Fomenko-Fuchs book on homotopical topology the another one definition of this map. It's similar to the above -- the only difference is the different way to choose the splitting $S^{m+n-1} = \left(D^{m} \times\right. \left.S^{n-1}\right) \cup\left(S^{m-1} \times D^{n}\right)$: picture. How to prove that both definitions are equivalent?
Radial projection from the boundary of $D^m\times D^n$ to $S^{m+n-1}$ give the second splitting but with the constant 1/2 possibly replaced by something else.
Please edit your question: first go to the point, then add the context after the statement as a complement. Please remove the typo and copy the text of the picture in the question, instead of just joining the picture.