(1)
Let \(F : M^n \to W^r\) and \(G : W^r \to V^s\) be smooth maps Let \(x\), \(y\), and \(z\) be local coordinates near \(p \in M^n\), \( F(p) \in W^r\), and \( G(F(p)) \in V^s\), respectively. Consider the composite map \(G \circ F : M \to V\),
(i) Using the coordinate basis \(\partial / \partial x\), we may investigate the differential \((G \circ F)_*\).
$$(G \circ F)_*\left(\frac{\partial}{\partial x^i}\right) = \sum_{k=1}^s \frac{\partial z^k}{\partial x^i} \frac{\partial}{\partial z^k}.$$
By the chain rule,
$$(G \circ F)_*\left(\frac{\partial}{\partial x^i}\right) = \sum_{k=1}^s \sum_{j=1}^r \frac{\partial z^k}{\partial y^j}\frac{\partial y^j}{\partial x^i}\frac{\partial}{\partial z^k}.$$
Since
$$F_*\left(\frac{\partial}{\partial x^i}\right) = \sum_{j=1}^r \frac{\partial y^j}{\partial x^i} \frac{\partial}{\partial y^j},$$
we see that
$$(G \circ F)_*\left(\frac{\partial}{\partial x^i}\right) = \sum_{k=1}^s \sum_{j=1}^r F_*^j\left(\frac{\partial}{\partial x^i}\right)\frac{\partial z^k}{\partial y^j}\frac{\partial}{\partial z^k},$$
and conclude
$$(G \circ F)_* = G_* \circ F_*.$$
(ii) Using the covector coordinate basis \(dx\), we may investigate the pull-back \((G \circ F)^*\).
$$(G \circ F)^*(dz^k) = \sum_{i=1}^n \frac{\partial z^k}{\partial x^i} dx^i.$$
By the chain rule,
$$(G \circ F)^*(dz^k) = \sum_{i=1}^n \sum_{j=1}^r \frac{\partial z^k}{\partial y^j} \frac{\partial y^j}{\partial x^i} dx^i.$$
Since
$$G^*(dz^k) = \sum_{j=1}^r \frac{\partial z^k}{\partial y^j} dy^j,$$
we see that
$$(G \circ F)^*(dz^k) = \sum_{i=1}^n \sum_{j=1}^r G_j^*(dz^k) \frac{\partial y^j}{\partial x^i} dx^i.$$
and conclude that
$$(G \circ F)^* = F^* \circ G^*.$$