= Solution
For based maps $\alpha:S^p\to Y$ and $\beta:S^q\to Y$, use the standard cell decomposition
$$
S^p\times S^q=(S^p\vee S^q)\cup_w D^{p+q}.
$$
The attaching map $w:S^{p+q-1}\to S^p\vee S^q$ is obtained from the boundary of $D^p\times D^q$, collapsing the appropriate coordinate boundaries. The geometric <Whitehead product> is the <homotopy class>
$$
[\alpha,\beta]_W=[(\alpha\vee\beta)\circ w]
\in\pi_{p+q-1}(Y).
$$
Orient the product cell consistently with the ordinary product orientation. Its class depends only on the two <homotopy> classes, and it is natural: $f_*[\alpha,\beta]_W=[f_*\alpha,f_*\beta]_W$.
For $p,q\ge2$ it is additive in both variables:
$$
[\alpha+\alpha',\beta]_W=[\alpha,\beta]_W+[\alpha',\beta]_W,
\qquad[\alpha,\beta+\beta']_W=[\alpha,\beta]_W+[\alpha,\beta']_W.
$$
Thus it is bilinear over $\mathbb Z$, and after <rationalization of a topological space> over $\mathbb Q$. In the geometric convention its symmetry is
$$
[\alpha,\beta]_W=(-1)^{pq}[\beta,\alpha]_W.
$$
For a genuine graded Lie convention, put a class in $\pi_pY$ in shifted degree $p-1$ and set $\{\alpha,\beta\}=(-1)^p[\alpha,\beta]_W$. Equivalently transport the <Samelson product> through <reduced suspension> and <loop space> adjunction. Writing $|\alpha|=p-1$, this bracket has degree zero, satisfies
$$
\{\alpha,\beta\}=-(-1)^{|\alpha||\beta|}\{\beta,\alpha\},
$$
and has the graded <Jacobi identity>
$$
(-1)^{|\alpha||\gamma|}\{\alpha,\{\beta,\gamma\}\}
+(-1)^{|\beta||\alpha|}\{\beta,\{\gamma,\alpha\}\}
+(-1)^{|\gamma||\beta|}\{\gamma,\{\alpha,\beta\}\}=0.
$$
These are two equivalent sign presentations of the same <Whitehead product>. In particular the square of an odd-dimensional geometric class is two-torsion, whereas an even-dimensional sphere class can have a nonzero rational square. <Reduced suspension> kills <Whitehead products>, since the suspended attaching map of the product cell is null-homotopic. No proofs of these remaining algebraic properties are needed here.
Bilinearity has a specific fundamental-group exception. If both inputs lie in $\pi_1Y$, the attaching loop of the torus is the <commutator>, so the operation is $ghg^{-1}h^{-1}$, with the possible inverse corresponding to reversal of its boundary orientation. This is not a homomorphism in either input in a general nonabelian group.
If one input is $g\in\pi_1Y$ and the other is $\beta\in\pi_qY$, $q\ge2$, the product records the action difference $g\cdot\beta-\beta$, up to the fixed overall boundary sign. Choose the action convention matching that boundary orientation and write $\delta_g\beta=g\cdot\beta-\beta$. Then
$$
\delta_g(\beta+\gamma)=\delta_g\beta+\delta_g\gamma,
\qquad
\delta_{gh}\beta=\delta_g\beta+g\cdot(\delta_h\beta).
$$
The second formula is a crossed-additivity law, not ordinary additivity in $g$. It follows by expanding $gh\cdot\beta-\beta=g\cdot(h\cdot\beta-\beta)+(g\cdot\beta-\beta)$. Consequently the higher-homotopy variable remains additive, while the fundamental-group variable is generally not. If the action is trivial the mixed product is zero. This explains exactly where the usual bilinearity statement fails.
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