Solution:
$(1)$ Since all terms are positive in the geometric sequence $\{a_n\}$ with the first three terms being $a-2$, $4$, $2a$,
$\therefore 4^2=2a(a-2)$, which simplifies to $a^2-2a-8=0$. Solving this, we get $a=4$ or $-2$.
Since $a > 0$, $\therefore a=4$.
$\therefore a_1=2$, $a_2=4$, and the common ratio $q= \dfrac {4}{2}=2$.
$\therefore S_k=62= \dfrac {2(2^k-1)}{2-1}$, solving this gives $k=5$.
$\therefore a=4$, $k=5$.
$(2)$ From $(1)$, we have: $a_n=2^n$.
$b_n=(2n-1)a_n=(2n-1)⋅2^n$.
$\therefore$ The sum of the first $n$ terms of the sequence $\{b_n\}$, $T_n=2+3×2^2+5×2^3+…+(2n-1)⋅2^n$,
$\therefore 2T_n=2^2+3×2^3+…+(2n-3)⋅2^n+(2n-1)⋅2^{n+1}$,
$\therefore -T_n=2+2(2^2+2^3+…+2^n)-(2n-1)⋅2^{n+1}= \dfrac {4(2^n-1)}{2-1}-2-(2n-1)⋅2^{n+1}=(3-2n)⋅2^{n+1}-6$,
$\therefore T_n=(2n-3)⋅2^{n+1}+6$.
Thus, for $(1)$, we have $a=4$ and $k=5$, so $\boxed{a=4, k=5}$.
For $(2)$, the sum of the first $n$ terms of the sequence $\{b_n\}$ is $\boxed{T_n=(2n-3)⋅2^{n+1}+6}$.