Why is $P(t < T leq t + dt) = f(t)dt$?

This statement is not strictly true, just approximately true assuming that $dt$ is small, and f is continuous and differentiable. It becomes strictly true in the limit $dt rightarrow 0$.

The probability density function $f(t)$ is defined, $$ P(t < T le t + Delta t) = int_t^{t+Delta t} f(t') dt'. $$ The RHS can be rewritten by transforming the integration variable, and with a Taylor expansion of the integrand we get, $$ begin{split} P(t < T le t + Delta t) &= int_0^{Delta t} f(t + tau) dtau \ &= int_0^{Delta t} left[ f(t) + f'(t) tau + frac{1}{2} f''(t) tau^2 + dots right] dtau \ &approx int_0^{Delta t} f(t) d tau \ &= f(t) Delta t. end{split} $$ In the second step I just Taylor expanded $f(t + tau)$ about $t$. If $Delta t$ is sufficiently small, then all factors of $tau$ at order $1$ or greater can be ignored compared to $f(t).$ You can see that in the limit as $Delta t rightarrow 0$, the approximate equality becomes a true equality.

In the area of survival analysis we can define $F(t)$ - the distribution function - as the probability that item fails in time interval $(0,t]$. Then the probability density function $f(t)$ is defined in the following way:

$$ f(t) = frac{d}{dt} F(t) = lim_{dtto 0} frac{F(t+dt) - F(t)}{dt} = lim_{dtto 0} frac{P(t < T leq t+dt)}{dt} $$

This implies that for small $dt$:

$$ P(t < T leq t+dt) approx f(t) dt $$

Note that there is not equal sign, it is just approximation for small $dt$.