A rate has a specific definition of $\frac{\# \mbox{events}}{\# \mbox{person-years}}$. A risk on the other hand refers to a particular individual's risk of experiencing an outcome of interest, and it is risk which is intrinsically related to the hazard (instantaneous risk). The language the question uses is consistent with this understanding. If I had to change it, I would say, "The death rate for smokers is twice that of *non-smokers". They also failed to mention whether these were age adjusted rates or not.
To understand this a little more deeply, relative rates and relative risks are estimated with fundamentally different models.
If you wanted to formalize a rate, you can think of this as estimating:
$$E \left( \frac{\# \mbox{events}}{\# \mbox{person-years}} \right) =\frac{\sum_i Pr(Y_i < t_i)} {\sum_i t_i} $$
($Y_i$ is the death time and $t_i$ is the observation time for the $i$-th individual, note the times are considered fixed and not random!)
You'll recognize the numerator is a bunch of CDFs, or 1-survival functions, and the relationship with survival functions and hazards is well known.
So if you took a ratio of rates:
$$ 2 = E \left( \frac{ \# \mbox{smoker deaths} \times \# \mbox{non-smoker person-years}}{\# {non-smoker deaths} \times \# \mbox{smoker person years}} \right) = \frac{\sum_i t_i}{\sum_j t_j} \frac{\sum_j Pr(Y_j < t_j)}{\sum_i Pr(Y_i < t_i)}$$
$$ = \frac{\sum_i t_i}{\sum_j t_j} \frac{ n_j-\sum_jS(t_j)}{n_i-\sum_iS(t_i)}$$
Since it's self study, you should probably do the algebra and solve the remainder of the equation!
Let $X$ denote the time of death (or time of failure if you
prefer a less morbid description). Suppose that $X$ is a continuous random
variable whose density function $f(t)$ is nonzero only on
$(0,\infty)$. Now, notice that it must be the case that $f(t)$
decays away to $0$ as $t \to \infty$ because if $f(t)$ does not decay away
as stated, then
$\displaystyle \int_{-\infty}^\infty f(t)\,\mathrm dt = 1$ cannot hold.
Thus, your notion that $f(T)$ is the probability of death at time $T$
(actually, it is $f(T)\Delta t$ that is (approximately)
the probability of death in the short interval $(T, T+\Delta t]$
of length $\Delta t$) leads to implausible and
unbelievable conclusions such as
You are more likely to die within the next month when you are thirty
years old than when you are ninety-eight years old.
whenever $f(t)$ is such that $f(30) > f(98)$.
The reason why $f(T)$ (or $f(T)\Delta t$) is the "wrong" probability
to look at is that the value of $f(T)$ is of interest only
to those who are alive at age $T$ (and still mentally alert enough
to read stats.SE on a regular basis!) What ought to be looked at
is the probability of a $T$-year old dying within the next month,
that is,
\begin{align}P\{(X \in (T, T+\Delta t] \mid X \geq T\}
&= \frac{P\{\left(X \in (T, T+\Delta t]\right) \cap \left(X\geq T\right)\}}{P\{X\geq T\}} &
\\ \scriptstyle{ \text{ definition of conditional probability}}\\
&= \frac{P\{X \in (T, T+\Delta t]\}}{P\{X\geq T\}}\\
&= \frac{f(T)\Delta t}{1-F(T)}
& \\ \scriptstyle{
\text{because }X\text{ is a continuous rv}}
\end{align}
Choosing $\Delta t$ to be a fortnight, a week, a day, an hour, a minute,
etc. we come to the conclusion that the (instantaneous) hazard
rate for a $T$-year old is
$$h(T) = \frac{f(T)}{1-F(T)}$$
in the sense that the approximate probability of death in the
next femtosecond
$(\Delta t)$ of a $T$-year old is $\displaystyle
\frac{f(T)\Delta t}{1-F(T)}.$
Note that in contrast to the density $f(t)$ integrating to $1$, the
integral
$\displaystyle \int_0^\infty h(t)\, \mathrm dt$ must diverge. This is because the CDF $F(t)$ is related to the hazard rate through
$$F(t) = 1 - \exp\left(-\int_0^t h(\tau)\, \mathrm d\tau\right)$$
and since $\lim_{t\to \infty}F(t) = 1$, it must be that
$$\lim_{t\to \infty} \int_0^t h(\tau)\, \mathrm d\tau = \infty,$$ or stated more formally, the integral of the hazard rate must diverge: there is no potential divergence as a previous edit claimed.
Typical hazard rates are increasing functions of time, but
constant hazard rates (exponential lifetimes) are possible. Both of these kinds of hazard rates obviously have divergent integrals. A less common scenario (for those who believe that things improve with age, like fine wine does) is a hazard rate that decreases with time but slowly enough that
the integral diverges.
Best Answer
It is the expected number of times you are expected to experience the event per time interval given that you have survived thus far. The key difference with your definition is that it is a rate not a probability.