Conditional Distributions in Hierarchy of Probability Spaces Consider the probability model
$$
X \mid Y = y \sim Poisson(\lambda y), \qquad Y \sim N(\mu, \sigma^2)
$$
where $\lambda, \mu, \sigma > 0$.  This models the number of bugs $X$ found on a leaf with surface area $Y$, in which the expected number of bugs is directly proportional to the area of the leaf.
Now, in usual conditional probability, there is only one probability space $(\Omega, \mathcal{F}, P)$, and a conditional distribution $\Lambda_{X \mid \mathcal{G}}$ is defined on this space where $\mathcal{G}$ is a sub sigma algebra of $\mathcal{F}$.  
In the example above, we would have $\mathcal{G} = \sigma(Y)$, but it seems like there should really be two probability spaces: the "first one" in which we sample a leaf to obtain a $y$, and the "second one" in which we sample the number of bugs on that leaf.  The number of bugs is not completely determined by the outcome of $Y$ -- hence the Poisson distribution -- and so there is still "randomness left" having performed the first experiment of selecting a $Y$.  How should this extra randomness be incorporated, if there is only one probability space?
 A: I would recommend that you think about a slightly simpler example in the fully discrete case:  suppose we have a fair six-sided die numbered from $1$ to $6$.  The die is cast and the value it shows represents the number of times you are to flip a coin with probability of heads $p \in (0,1)$.  Then let $X$ represent the random number of heads obtained.  $Y$ is the number rolled on the die.
In this situation, we have a hierarchical binomial/discrete uniform model; i.e. $$X \mid Y \sim \operatorname{Binomial}(Y,p), \quad Y \sim \operatorname{DiscreteUniform}(1,6).$$  (Note we have avoided any undefined outcomes, unlike the example that you provided.)  Now write out the probability space for this entire experiment, which is (reasonably) tractable, by specifying the sample space $\Omega$, the event space $\mathcal F$, and the probability measure $P$.   Where does the hierarchy come in?  Can you see now why defining this one space adequately characterizes the entire hierarchical model?
Another thing to think about is that if you look at your example, are there really only two spaces if you propose to count them the way you do, or would there actually be a countably infinite number, one corresponding to each outcome of $Y$?  Can you see why this is unnecessary?  A space can be defined for the entire experiment, not just its constituent processes; and as Did's comment observes, more than one random variable can be defined on a single space.
A: In a real-life application, I suppose it is taken for granted that
$\mu > a\sigma,$ where $a$ is at least four and maybe larger. This
makes $P(Y \leq 0)$ negligibly small. Otherwise, we can use the 'truncated
normal distribution' where the density $f_Y$ has positive values only in $(0, \infty)$ ($0$ otherwise), but is 'inflated' so that $\int_o^\infty f_Y(y)\,dy = 1.$
The PDF of $X$ can be simulated in R statistical software as follows. Strictly speaking, in order to
avoid error messages, the simulation
code has to take into account the (remote) possibility a value of $Y$ that is
$0$ or below. Also, the function rpois that generates random observations
from a Poisson distribution returns $0$ if the rate is $0$. The simulation
below uses $\mu = 100,\, \sigma=5,$ and $\lambda = 4.$ (Results should be
correct to several significant digits.)
m = 10^6;  mu = 100;  sg = 5;  lam = 4
y = rnorm(m, mu, sg)   # areas of a 10^6 leaves
y[y < 0] = 0           # to avoid error messages, set 'area'=0, if < 0
x = rpois(m, lam*y)    # number of bugs on each of 10^6 leaves
mean(x);  var(x)
## 399.9766            # aprx E(X) = 400
## 801.1809            # aprx Var(X)

It is not surprising that $E(X) = \lambda\mu = 400,$ as is well approximated
by the simulation. It is up to you to derive the value of $Var(X).$ (Maybe look under 'a random number of random variables'.)
Below is a histogram of the approximate distribution of $X$ based on
numbers of bugs on a million simulated leaves. The curve is the
normal density that matches $\hat \mu = 399.98$ and $\hat \sigma^2 = 801.18.$
The fit is reasonably good.

You give no clue about the course this is from or your background,
so I am not sure what approach is expected of you. The simplest task
would be to assume that $Y$ is nearly normal and to find its mean $\mu$ and
variance $\sigma^2.$ A deeper analytic treatment would be to
show that $X$ is approximately normal (perhaps dealing with the truncation of $Y$).
Sometimes discrete distributions that arise in this way are called 'overdispersed'
Poisson distributions: such a distribution has a variance that is numerically greater than its mean.
