Recall that for $\forall x\in\mathbb N$, $\forall m,u,v\in\mathbb N^*$, it holds ${(x^u\bmod m)}^n\equiv{(x^u)}^v\equiv x^{u\times v}\pmod m$, where $y\equiv x\pmod m$ means $m$ divides $x-y$, and $x\bmod m$ is the uniquely defined integer $y$ such that $0\le y<m$ and $y\equiv x\pmod m$.
The shared secret is $K=(g^a\bmod p)^b\bmod p=(g^b\bmod p)^a\bmod p$, or equivalently $K=g^{a\times b}\bmod p$. We are tasked to evaluate this for $a=6$, $b=8$, $g=2$, $p=19$.
The method in the question goes:
$$\begin{array}{}
K&={(2^6\bmod19)}^8\bmod19&&=2^{6\times8}\bmod19\\
&=2^{(3\times2)\times8}\bmod19&={(2^3)}^{2\times8}\bmod19&=8^{2\times8}\bmod19\\
&={(8^2)}^8\bmod19&=64^8\bmod19\\
&&=(64-19\times3)^8\bmod19&=7^8\bmod19\\
&={(7^2)}^4\bmod19&=49^4\bmod19\\
&&=(49-19\times2)^4\bmod19&=11^4\bmod19\\
&={(11^2)}^2\bmod19&=121^2\bmod19\\
&&=(121-19\times6)^2\bmod19&=7^2\bmod19\\
&=49\bmod19&=49-19\times2&=11\\
\end{array}$$
and that (keeping the rightmost column) can be condensed to:
$$K\equiv2^{6\times8}\equiv8^{2\times8}\equiv7^8\equiv11^4\equiv7^2\equiv11\pmod{19}\ \text{ thus }\ K=11$$
If I was to compute this without calculator, I would write this as $K=2^{48}\bmod19$, then use Fermat's Little Theorem. It says that when $p$ is prime and $g$ is not a multiple of $p$, if holds $g^{p-1}\bmod p=1$. That allows to reduce modulo $(p-1)$ any exponent of $g$ when computing modulo $p$. The complete calculation goes:
$$\begin{array}{}
K&=2^{6\times8}\bmod19&&=2^{48}\bmod19\\
&=2^{48\bmod(19-1)}\bmod19&=2^{48-18\times2}\bmod19&=2^{12}\bmod19\\
&=4096\bmod19&=4096-19\times215&=11\end{array}$$
Note: In $48-18\times2=12$, the $2$ is obtained as the quotient $\lfloor48/18\rfloor$, much like in $4096-19\times215=11$ the $215$ is $\lfloor4096/19\rfloor$.
Actual cryptography uses integers much too large for reliable human computation; e.g. $p$ could be 2048-bit, that is 617 decimal digits.
