You only get a necessary condition for $k$ being a solution.
Suppose you have
\begin{cases}
3x=2 \\
7x=2
\end{cases}
If you subtract, you get $4x=0$, which implies $x=0$. Does this mean $x=0$ is a solution? Of course not, because $1$ satisfies neither equation. What you can say is “if $x$ is a solution, then $x=0$”.
But if you multiply the first equation by $2$ and subtract from the second, you find $x=-2$. So “if $x$ is a solution, then $x=-2$”, which is obviously a contradiction with the previous finding. Assuming a solution exists leads to a contradiction, so the system has no solution.
In your case you could indeed multiply the first congruence by $2$ and subtract from the second, getting
$$
k\equiv -2\pmod{8}
$$
Is this a solution? Let's check: if $k\equiv-2$, then
$$
3k\equiv-6\equiv2
\qquad
7k\equiv-14\equiv2
\qquad
\pmod{8}
$$
Hurray! The solution is good! Then $k=8q-2$ (or, equivalently, $k=8q+6$) are all the integer solutions.
In a different fashion, you can note that $3\cdot3\equiv1\pmod{8}$, so the first equation is equivalent to
$$
3\cdot3k\equiv3\cdot2\pmod{8}
$$
that is,
$$
k\equiv6\pmod{8}
$$
Since $-7\equiv1\pmod{8}$, the second equation is equivalent to
$$
-7k\equiv-2\pmod{8}
$$
that is,
$$
k\equiv-2\equiv6\pmod{8}
$$
Thus the two equations are exactly the same.
