The integral expression for $Y(t)$ refers to some stochastic calculus
involving the Poisson process. However it can be understood as a sum
of random infinitesimal contributions and handled as such on an intuitive
ground provided that the assumptions are carefully understood and
checked in the derivation.
Note that for the specific problem discussed here an equivalent representation
for $Y(t)$ relies on the infinite sequences of r.vs: the Poissonian event times
$T_i$, the related pulses depth $X_i$ and pulse lengths $L_i$
$$
Y(t) = \sum_i X_i \, 1_{\{T_i < t < T_i + L_i\}}.
$$
This expression can be used to find the expectation
$$
\mathbb{E}\{Y(t)\} = \sum_i \mathbb{E}[X_i]\,
\mathbb{E}[1_{\{T_i < t < T_i + L_i\}}] = \mathbb{E}[X]
\, \sum_i \mathbb{E}[1_{\{T_i < t < T_i + L_i\}}].
$$
where the independence of $X_i$ and $[T_i,\,L_i]$ has been used in the
first step. Since $T_i$ and $L_i$ are independent, we have
$\mathbb{E}[1_{\{T_i < t < T_i + L_i\}}] = \mathbb{E}[1_{\{T_i < t <
T_i + L\}}]$. So the sum is the expectation of the number of
events $T_i$ falling in the random interval $(t - L, \, t)$, which
unsurprisingly can be shown to be $\lambda \mu_L$.
Coming back to the integral expression, we can use the equivalent expression
$$
Y(t) = \int_{s = -\infty}^t X_s(t-s) \,\text{d}N_s
$$
By exchanging the expectation and integral
$$
\mathbb{E}\{Y(t)\} = \int_{s =-\infty}^t
\mathbb{E}\{X_{s}(t - s) \,\text{d}N_{s}\} = \int_{s =-\infty}^t
\mathbb{E}\{X_{s}(t - s)\} \, \lambda \text{d}s.
$$
To justify the second expression, the expectation in the integral is a
product of expectations due to the independence of the sequence
$[X_i,\,L_i]$ of pulse variables and the sequence $T_i$ of events;
moreover $\mathbb{E}\{\text{d}N_{s}\} = \lambda \text{d}s$ is the
expected increment of $N_s$ on the time interval $(s, \,
s+\text{d}s)$. Now using the distribution of the r.v. $X_{s}(t - s)$
(with mixed type) as provided in (2.4) and the independence of $X$ and
$L$, we get $\mathbb{E}\{X_{s}(t - s)\} = \mathbb{E}[X]
\,\mathcal{F}_L(t-s)$. The result then comes from $\int_{u=0}^\infty
\mathcal{F}_L(u) \text{d}u = \mathbb{E}[L]$.
The formula for the autocorrelation can be derived by writing the product
$Y(t)Y(t + \tau)$ as a double integral and then taking the expectation.
We can use the rule
$$
\mathbb{E}[\text{d}N_s \text{d}N_{s'}] =
\begin{cases}
\lambda \, \text{d}s & s= s',\\
0 & s \neq s'
\end{cases}
$$
which relates to the independent increments property of $N_t$.
EDIT: As highlighted by comments by @lmnop, the final part of my answer
concerning the autocovariance is misleading. See section 9.6 of D.R. Cox
and H.D. Miller (1965) The Theory of Stochastic Processes about the linear
filtering of a Point process including the Poisson case. The simplest
case concerns the linear filtering a Poisson process
$$
Y(t) = \int_{s= -\infty}^{t} g(t- s) \, \text{d}N_s
$$
where $g(t)$ is a deterministic function. The covariance of $Y(t)$ and
$Y(t+\tau)$ can be obtained by writing a double integral, using the
bilinearity of the covariance and
$$ \text{Cov}\{\text{d}N_s,\, \text{d}N_{s'}\} = \begin{cases} \lambda
\, \text{d}s & s= s',\\ 0 & s \neq s'. \end{cases}
$$
We get
$$ \text{Cov}\{Y(t),\, Y(t+ \tau)\} = \lambda \, \int_{s= -\infty}^t g(t -s) g(t + \tau -s)
\,\text{d}s = \lambda \int_{0}^\infty g(u) g(u + \tau)\,\text{d}u
$$
Under some assumptions which hold in the application paper cited in the
OP, this extends to the case where $g(t)$ is a stochastic process
related to the Poisson Process $N_t$, but then we must then use an
expectation in the integral, as in
$$
\text{Cov}\{Y(t),\, Y(t+ \tau)\} = \lambda
\int_{s= -\infty}^t \mathbb{E}\{g(t-s) g(t+\tau-s)\}\,\text{d}s.
$$
See the book for a derivation based on infinitesimal increments. I
guess that some more recent references could also be found, in
relation to the formal rule $\text{d}N_s \text{d}N_s =
\text{d}N_s$. Moreover the autocovariance can be found using discrete
sums as done here for the expectation.
