Why is $\int_0^t t \, dW_s$ not a martingale?

The process $(t W_t)_{t \geq 0}$ is indeed not a martingale. The reason is simply that you cannot use the martingale property of stochastic integrals to deduce the martingale property of the process. Correctly stated the martingale property of stochastic integrals driven by Brownian motion reads as follows:

Theorem Let $f:\Omega \times [0,\infty) \to \mathbb{R}$ be a progressively measurable function such that $\mathbb{E}\left( \int_0^t f(s)^2 \,ds \right)<\infty$. Then the process $$M_t := \int_0^t f(s) \, dW_s$$ is a martingale.

Note that $f$ is not allowed to depend on $t$, i.e. the statement does not claim that

$$N_t :=\int_0^t f(s,t) \, dW_s$$

is a martingale for any progressively measurable and square integrable function $f$.

If we choose $f(s) := s$, then the above statement yields that

$$M_t = \int_0^t s \, dW_s$$

is a martingale - which is true.

We can also choose $f(s)=t_0$ for some fixed $t_0$; then we find that

$$M_t = t_0 W_t$$

is a martingale - which is also true (because $t_0$ is a fixed constant).