For any vector $y\in\mathbb F_2^n$, use its <Hamming weight> $w(y)=\#\{j:y_j=1\}$. The stated product identity also follows directly by summing each coordinate independently:
$$
\sum_y u^{w(y)}(-1)^{x\cdot y}
=\prod_{j=1}^n(1+(-1)^{x_j}u)
=(1-u)^{w(x)}(1+u)^{n-w(x)}.
$$
For $t\ne0$, interchange the two finite sums with $u=s/t$. <Character orthogonality> evaluates the inner sum over codewords, whereas the product identity evaluates the sum over all binary vectors. Thus
$$
2^k\sum_{y\in C^\perp}(s/t)^{w(y)}
=\sum_{x\in C}(1-s/t)^{w(x)}(1+s/t)^{n-w(x)}.
$$
Multiply by $t^n$ to obtain the <MacWilliams identity> in the paper's weight-enumerator convention:
$$
\boxed{W_{C^\perp}(s,t)=2^{-k}W_C(t-s,t+s)}.
$$
Both sides are <polynomials>, so equality for $t\ne0$ extends to $t=0$ as well; the intermediate quotient is not a restriction on the final identity.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 1 / 10G / ii / Solution

{c}

For a binary <linear code> of dimension $k$ and its <dual code>, the <weight enumerator> convention $s^{\operatorname{wt}(x)}t^{n-\operatorname{wt}(x)}$ gives
$$
W_{C^\perp}(s,t)=2^{-k}W_C(t-s,t+s).
$$
To prove it, insert $\mathbf1_{C^\perp}(y)=2^{-k}\sum_{x\in C}(-1)^{x\cdot y}$ into the enumerator sum. Summing independently over each coordinate of $y$ yields a factor $t+s$ if $x_j=0$ and $t-s$ if $x_j=1$.

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 Incoming links: MacWilliams identity