Help with JK flipflop counter?

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First, you don't need JK FF 0, as you have observed already Q0 is always 0 whatever the conditions are. Why put something that does nothing in a circuit? Just ground the Q0 and you are set.

Second, notice that if you have a truth table of the counter and you cover the Q0 column you are actually looking at an ordinary 2-bit up and down counter's truth table. So basically, counting 0, 2, 4, 6, 0 is equal to 2*0, 2*1, 2*2, 2*3, 2*0. This is because multiplying by two in binary domain is equal to shifting left by one bit.

So, just design a 2-bit up/down counter, shift it left by one bit (by renaming the index to index + 1) and then add a grounded Q0.