Units of s represent the Laplace transform while units of z represent the z transform. It is often much easier to transform a time domain (units of t) signal to Laplace given that it translates integro-differential equations into simple algebra. The z transform does the same thing, but takes advantage of the repetition from sampling to make the transform even easier.
"s" is any complex variable (with real and imaginary values), usually expressed $s=\sigma + j\omega$. When you limit s to only be imaginary ($s=j\omega$), the result is the Fourier Transform, which is a single-valued function versus frequency. Thus for a system, $H(s)$ would be the general transfer function of s and $H(j\omega)$ is the frequency response.
Similary "z" is any complex variable, but due to the mapping from analog to digital in the construction of the z -transform, what is the frequency axis in the s-plane (the $j\omega$ axis) becomes the unit circle in the z plane ($e^{j\omega}$), where $\omega$ here is typically normalized radian frequency (see What is normalized frequency) with a unique range of $0$ to $2\pi$. Thus a digital system would have a general transfer function given as $H(z)$ while the frequency response would be $H(e^{j\omega}$). Conveniently, the z transform of one sample delay is simply $z^{-1}$, which hints at the great utility of using the z-transform for discrete time systems rather than the Laplace transform (which you could certainly still do instead if you want to take a mathematical beating!).
(Note: The mapping from $j\omega$ to the unit circle applies to many but not all mapping techniques: see z-Transform Methods: Definition vs. Integration Rule)
That said, to model the complete system digitally as a discrete system, you would translate the blocks that are functions of s to be functions of z. It is the process of translating, or mapping, from s to z where the sampling rate comes into play. There are several mapping techniques to choose from and a few popular ones are the matched-z transform, the method of impulse invariance and the bilinear transform. (For more on that see How/why are the $\mathcal Z$-transform and unit delays related? ) A simple example of this is the mapping of integration and differentiation operations from s to z using the matched-z transform (which by the way is identical to the method of impulse invariance for an all pole system):
Integration in time is equivalent to multiplying by $\frac{1}{s}$ in Laplace (there we see the conversion to simple algebra!). This is mapped to the following in the z domain:
$$\frac{T}{1-z^{-1}}$$
Differentiation in time is equivalent to multiplying by $s$ in Laplace (along with subtracting the initial condition which I didn't include). This is mapped to the following in the z-domain:
$$\frac{1-z^{-1}}{T}$$
where $T$ represents the time of one sample period.
Notice the factor of $T$ which is the sample period. The differentiation example is very easy to see what is going on: $1-z^{-1}$ is a simple difference of two successive samples:
$$H(z) = 1 - z^{-1}$$
$$\frac{Y(z)}{X(z)} = 1 - z^{-1}$$
$$Y(z) = X(z) - X(z)z^{-1}$$
Which is a (not the only) discrete approximation of a continuous time domain derivative. Consider taking the derivative of any time domain function: as we sample at higher and higher rates, the difference between two successive samples will get smaller, thus by dividing by the sample duration for the case of the derivative which is also getting smaller, we normalize the result to be independent of the sampling rate.
In your case specifically, this would occur in the mapping from s to z for the continuous time transfer function given by $H_b(s)$.
