Let $\, n = d_0 + d_1 10 + d_2 10^2 + \cdots + d_k 10^k = P(10),\,$ where $P(n)$ is $\,n$'s $ $ radix $10$ polynomial, with digits $\,d_i.\,$ You are evaluating $\ \color{#0a0}{2\, P(3)}\,$ since your sequence $\, 2, 6, 4, 5,\ldots \equiv 2\cdot 3^k\pmod{\!7}$.
Now $\,{\rm mod}\ 7\!:\ \color{#c00}{10\equiv 3}\,$ $\Rightarrow$ $\,P(\color{#c00}{10})\equiv P(\color{#c00}3),\,$ by the Polynomial Congruence Rule, therefore this congruence implies
$\,7\mid P(10)\iff 7\mid P(3)\!\iff\! 7\mid \color{#0a0}{2P(3)},\,$ using divisibility mod reduction for the first $(\!\!\iff\!\!);$ whereas the second arrow uses by $\color{#0a0}{(7,2)=1}$ and Euclid's Lemma.
So your method yields a valid divisibility test. If we instead used sequence $\,3^k \equiv 1,3,2,6,4,5\,$ then we'd obtain the standard test $\,7\mid P(10)\iff 7\mid P(3),\,$ which is usually optimized by evaluating $\,P(3)\,$ modulo $7,\,$ i.e. use mod $7$ arithmetic when we compute the "dot product". This method has a big advantage: $\,P(3)\,$ (but not $\,2P(3))\,$ has the same remainder mod $7$ as $\,P(10)\,$ so we can use it to do arithmetic mod $7$, e.g. as a check of decimal arithmetic - just like casting out nines uses $\,{\rm mod}\ 9\!:\ 10\equiv 1\,$ $\Rightarrow$ $\,P(10)\equiv P(1),\,$ by the Polynomial Congruence Rule.
Remark $\ $ There's a better way - a universal divisibility test that is simpler and much easier recalled, viz. evaluate the above radix polynomial $\,P(x)\,$ in nested (Horner) form, using modular arithmetic. For example, consider evaluating a $3$ digit radix $10$ number modulo $7.\,$ In Horner form $\rm\ d_2\ d_1\ d_0 \ $ is $\rm\: (d_2\cdot \color{#c00}{10} + d_1)\ \color{#c00}{10} + d_0\, \equiv\ (d_2\cdot\color{#c00} 3 + d_1)\ \color{#c00} 3 + d_0\pmod 7\ $ by $\rm\ \color{#c00}{10\equiv 3}\pmod{\! 7}.\,$ Thus we compute the remainder $\rm\!\! \mod {\!7}\, $ as follows: start with the leading digit then repeatedly apply the operation: $ $ multiply by $3$ then add the next digit (doing all of the arithmetic $\!\!\mod{\! 7}$
For example, let's use this algorithm to reduce $\rm\ 43211\ \:(mod\ 7)\:.\:$ The algorithm consists of repeatedly replacing the first two leading digits $\rm\ d_n\ d_{n-1}\ $ by $\rm\ \bbox[5px,border:1px solid #c00]{3 \,d_n + d_{n-1}\bmod 7}\ $ viz.
$$\begin{array}{rrl} &\color{#0A0}{4\ 3}\ 2\ 1\ 1&\\
\equiv\!\!\!\! &\color{#c00}{1\ 2}\ 1\ 1 &\!{\rm by}\, \ \bbox[5px,border:1px solid #c00]{3\cdot \color{#0a0}4 +
\color{#0a0}3} \equiv \color{#c00}1\\
\equiv\!\!\!\! &\color{#0af}{5\ 1}\ 1&\!{\rm by}\ \ \ 3\cdot \color{#c00}1 +
\color{#c00}2\ \equiv\ \color{#0af}5\\
\equiv\!\!\!\! & \color{#f60}{2\ 1}&\!{\rm by}\ \ \ 3\cdot \color{#0af}5 +
\color{#0af}1\ \equiv\ \color{#f60}2\\
\equiv\!\!\!\! &\color{#8d0}0&\!{\rm by}\ \ \ 3\cdot \color{#f60}2 +
\color{#f60}1\ \equiv\ \color{#8d0}0
\end{array}\qquad\qquad$$
Hence $\rm\ 43211\equiv 0\:\ (mod\ 7),\:$ indeed $\rm\ 43211 = 7\cdot 6173.\:$ Generally the modular arithmetic is simpler if one uses a balanced system of representatives, e.g. $\rm\: \pm\{0,1,2,3\}\ \:(mod\ 7),\,$ e.g. see here. Notice that for modulus $11$ or $9\:$ the above method reduces to the well-known divisibility tests by $11$ or $9\:$ (a.k.a. casting out nines or elevens).
The above is much better than a divisibility test since it actually calculates the remainder mod $7$ (unlike the above divisibility test, which can only be used to test if the remainder $= 0).\,$ Thus - as in casting out nines - we can perform further arithmetic with the remainders, e.g. using them to help check the correctness of calculations.
