Multiply a measured quantity of a control-system in the frequency domain during continious operation
I assume that the transfer function that you are using is of a similar form as $V(\eta)$ in the image of your question, with $\eta = \Delta f / f_0$. Often a transfer function is written in terms of the complex Laplace variable $s$, but I assume that your $\eta$ represents some sort of normalized frequency. That is equivalent to using $s = i\,k\,\omega$ with $i$ the unit imaginary number, $k$ some normalization constant and $\omega$ the angular frequency. If you rewrite your transfer function in terms of $s$ instead of $\eta$ you can use one of the many known discretization methods. A simple but good discretization method is the bilinear transform, which requires you to substitute $s$ with the following expression in the transfer function

$$
s = \frac{2}{T} \frac{z - 1}{z + 1}, \tag{1}
$$

with $T$ the sample time of your sensor. After simplifying the resulting (discrete time) transfer function you should have something of the form

$$
V(z) = \frac{b_m\,z^m + b_{m-1}\,z^{m-1} + \cdots + b_0}{z^n + c_{n-1}\,z^{n-1} + \cdots + c_0}, \tag{2}
$$

with $n \geq m$ and all $b_i$ and $c_i$ real scalar constants. When multiplying a discrete signal with the variable $z$ can be interpreted as shifting that signal one sample in time. So given the input signal $a[k]$ (with $k$ an integer used as an index and the time associated with that index could for example be $t = k\,T$) the output $F[k]$ can be related to each other using $(2)$

\begin{align}
(z^n + c_{n-1}\,z^{n-1} + \cdots + c_0)\,F[k] &amp;= (b_m\,z^m + b_{m-1}\,z^{m-1} + \cdots + b_0)\,a[k], \\
F[k+n] + c_{n-1}\,F[k+n-1] + \cdots + c_0\,F[k] &amp;= b_m\,a[k+m] + b_{m-1}\,a[k+m-1] + \cdots + b_0\,a[k].
\end{align}

Solving this for $F[k]$ gives

$$
F[k] = b_m\,a[k+m-n] + b_{m-1}\,a[k+m-1-n] + \cdots + b_0\,a[k-n] - c_{n-1}\,F[k-1] - \cdots - c_0\,F[k-n],
$$

so a weighted sum of previous inputs and outputs. This relation for $F[k]$ allows you to calculate a new output every time you get a new input, so no need to calculate it using chunks. I can be noted that initially you do not know the past values of the input and output. However, usually one set those to zero or some other initial value, which should only change the initial transient behavior of the output (assuming that the filter is stable).