Convolving a discrete time sequence with a continuous-time filter?
Conceptually, if used as an input to a continuous-time system, a discrete-time signal is represented as a weighted sum of Dirac delta impulses (as pointed out in a comment by MBaz):

$$g(t)=\sum_kg[k]\delta(t-kT)\tag{1}$$

If $g(t)$ in $(1)$ is the input to a continuous-time LTI system with impulse response $h(t)$, the output is given by

$$y(t)=\sum_kg[k]h(t-kT)\tag{2}$$

This expression is similar, but not identical, to the one you got. In your case, the impulse response $h(t)$ is simply the convolution of $h_1(t)$ and $h_2(t)$:

$$h(t)=(h_1\star h_2)(t)\tag{3}$$

Sampling $y(t)$ with a sampling interval $T$ means to replace the continuous time variable $t$ by $nT$ in $(2)$:

$$y[n]=y(nT)=\sum_kg[k]h((n-k)T)\tag{4}$$

If we define a discrete-time impulse response $h[n]=h(nT)$, we can write $(4)$ as a discrete-time convolution:

$$y[n]=(g\star h)[n]\tag{5}$$

Now we just have to filter $y[n]$ with the discrete-time impulse response $h_3[n]$ to obtain the final output signal:

$$\begin{align}z[n]&amp;=(y\star h_3)[n]\\&amp;=(g\star h\star h_3)[n]\end{align}\tag{6}$$

with

$$h[n]=h(nT)=(h_1\star h_2)(nT)=\int_{-\infty}^{\infty}h_1(\tau)h_2(nT-\tau)d\tau\tag{7}$$