Please explain this formal definition of computation
We are missing some context so I have no idea what point Knuth is trying to make, but here's how to interpret a Turing machine this way. Perhaps it will help you understand what's going on. In general, a good way of getting a grip on a concept is playing with it. In the case of programming paradigms, that means writing a program. In this case, I will show how to write any program.

Suppose the tape of the Turing machine has symbols $\{0,1,\epsilon\}$ (where $\epsilon$ stands for "empty"), and add one more symbol which represents the location of the head $H$. Your states are going to be pairs of the form $(q,\alpha)$, where $q$ is a state of the Turing machine, and $\alpha \in \{0,\ldots,14\}$. We also identify $(F,0)$ with $N$ for any final state.

On (non-empty) input $x$, your starting point will be $(Hx,(s,0))$, where $s$ is the starting state. The difficult part is to encode states. Suppose that at state $q$, upon reading input $x$, you replace it with $a(q,x)$, move in direction $D(q,x) \in \{L,R\}$, and switch to state $\sigma(q,x)$. For the $\theta$s, we have
$$
\begin{align*}
\theta_{q,0} &amp;= 0H0, &amp; \theta_{q,1} &amp;= 0H1, &amp; \theta_{q,2} &amp;= 0H\epsilon, \\
\theta_{q,3} &amp;= 1H0, &amp; \theta_{q,4} &amp;= 1H1, &amp; \theta_{q,5} &amp;= 1H\epsilon, \\
\theta_{q,6} &amp;= \epsilon H0 &amp; \theta_{q,7} &amp;= \epsilon H1, &amp; \theta_{q,8} &amp;= \epsilon H\epsilon, \\
\theta_{q,9} &amp;= H0, &amp; \theta_{q,10} &amp;= H1, &amp; \theta_{q,11} &amp;= H\epsilon, \\
\theta_{q,12} &amp;= 0H, &amp; \theta_{q,13} &amp;= 1H, &amp; \theta_{q,14} &amp;= \epsilon H.
\end{align*}
$$
For the $a$s, we have $a_{q,i} = (q,i+1)$ for $i &lt; 14$, and $a_{q,14} = (q,14)$, though we should never really get that far. For the $b$s, we have
$$
\begin{align*}
&amp;b_{q,0} = b_{q,3} = b_{q,6} = b_{q,9} = (\sigma(q,0),0), \\
&amp;b_{q,1} = b_{q,4} = b_{q,7} = b_{q,10} = (\sigma(q,1),0), \\
&amp;b_{q,2} = b_{q,5} = b_{q,8} = b_{q,11} = b_{q,12} = b_{q,13} = b_{q,14} = (\sigma(q,\epsilon),0).
\end{align*}
$$
Now it remains to determine the $\psi$s. Let $a_0 = a(q,0)$. If $D(q,0) = L$ then
$$
\begin{align*}
\psi_{q,0} &amp;= H0a_0, &amp; \psi_{q,3} &amp;= H1a_0, &amp; \psi_{q,6} &amp;= \psi_{q,9} = H\epsilon a_0.
\end{align*}
$$
If $D(q,0) = R$ then
$$
\begin{align*}
\psi_{q,0} &amp;= 0a_0H, &amp; \psi_{q,3} &amp;= 1a_0H, &amp; \psi_{q,6} &amp;= \epsilon a_0 H, &amp; \psi_{q,9} &amp;= a_0H\epsilon.
\end{align*}
$$
Next, let $a_1 = a(q,1)$. If $D(q,1) = L$ then
$$
\begin{align*}
\psi_{q,1} &amp;= H0a_1, &amp; \psi_{q,4} &amp;= H1a_1, &amp; \psi_{q,7} &amp;= \psi_{q,10} = H\epsilon a_1.
\end{align*}
$$
If $D(q,1) = R$ then
$$
\begin{align*}
\psi_{q,1} &amp;= 0a_1H, &amp; \psi_{q,4} &amp;= 1a_1H, &amp; \psi_{q,7} &amp;= \epsilon a_1 H, &amp; \psi_{q,10} &amp;= a_1 H\epsilon.
\end{align*}
$$
Finally, let $a_\epsilon = a(q,\epsilon)$. If $D(q,\epsilon) = L$ then
$$
\begin{align*}
\psi_{q,2} &amp;= H0a_\epsilon, &amp; \psi_{q,5} &amp;= H1a_\epsilon, &amp; \psi_{q,8} &amp;= \psi_{q,11} = H\epsilon a_\epsilon, \\
\psi_{q,12} &amp;= H0a_\epsilon, &amp; \psi_{q,13} &amp;= H1a_\epsilon, &amp;\psi_{q,14} &amp;= H\epsilon a_\epsilon.
\end{align*}
$$
If $D(q,\epsilon) = R$ then
$$
\begin{align*}
\psi_{q,2} &amp;= 0a_\epsilon H, &amp; \psi_{q,5} &amp;= 1a_\epsilon H, &amp; \psi_{q,8} &amp;= \epsilon a_\epsilon H, &amp; \psi_{q,11} &amp;= a_\epsilon H\epsilon, \\
\psi_{q,12} &amp;= 0a_\epsilon H, &amp; \psi_{q,13} &amp;= 1a_\epsilon H, &amp; \psi_{q,14} &amp;= \epsilon a_\epsilon H.
\end{align*}
$$

Now apply $f$ repeatedly until you get stuck. If you follow the construction, you will see that we have simulated the running of the Turing machine.