Here is the question:
Show that is natural in both variables. That is suppose $f: X \rightarrow Y, u \in \tilde{H^{*}}(Y), \alpha \in \tilde{H_{*}}(X).$ Then we can form the cap products$$ \langle u, f_{*}(\alpha)\rangle \in \tilde{H}_{n-k}(Y)$$ and $$\langle f^{*}(u), \alpha)\rangle \in \tilde{H}_{n-k}(X).$$
Show that $$f_{*}(\langle f^{*}(u), \alpha) \rangle) = \langle u, f_{*}(\alpha) \rangle. $$
Could anyone help me in proving this exercise please?
There is a chain map $C_*(A ) \rightarrow C_*(A) \otimes C_*(A)$ called the Alexander-Whitney chain map which is chain homotopic to $C_*(A) \rightarrow C_*(A \times A) \rightarrow C_*(A) \otimes C_*(A)$ where the first map is the diagonal map and the second map is the Künneth map.
Then for a $p-$chain, $\sigma \in C_p(X;R)$ and a $q-$cochain, $\phi \in C^q(X;R)$ we see using this that $\sigma \cap \phi = \phi(_q\sigma) * \sigma_{p-q}$ where $_i \sigma$ and $\sigma_j$ are the $i-$th back face and $j-$th front face respectively.
To understand this chain map and front/back faces better I recommend reading the relevant part of the chapter titled "Multiplication" in James V. Wicks "Homology theory".
Changing your notation a bit, if we have $u$ and $\alpha$ be representatives of your cohomology classes $[u]$ and $[\alpha]$ and $f_\#:C_*(X;R) \rightarrow C_*(Y;R)$ and $f^\#: C^*(Y;R) \rightarrow C^*(X;R)$ are the induced chain maps by $f$ and $r * \eta \in C_*(A;R)$ just denotes multiplication of $\eta \in C_*(A)$ and $r \in R$ we see that
$f_\#(\alpha \cap f^\#(u)) = f_\#(\alpha \cap (u \circ f_\#))$
$= f_\#((u \circ f_\#)(_n\alpha) * \alpha_{n-k}) = (u \circ f_\#)(_n\alpha) * f_\#(\alpha_{n-k}) = (u( f_\#(_n\alpha)) * f_\#(\alpha)_{n-k}$
$= (u( _nf_\#(\alpha)) * f_\#(\alpha)_{n-k} = f_\#(\alpha) \cap u$.
First equality comes from the fact that $f^\#(u) = u \circ f_\#$.
The second equality uses the identity $\sigma \cap \phi = \phi(_q\sigma) * \sigma_{p-q}$ for $\sigma = \alpha$ and $\phi = u \circ f_\#$.
Third equality uses that $f_\#$ is $R-$linear and $(u \circ f_\#)(_n \alpha) \in R$.
Fourth and fifth equality uses naturality of taking back and front faces of simplices which says that $(-)_{n-k}$ and $_n(-)$ commute with chain maps $f_\#$.
The final equality uses the formula $\sigma \cap \phi = \phi(_q\sigma) \sigma_{p-q}$ once more for $\sigma = f_\#(\alpha)$ and $\phi = u$.