Designing a special 4-bit counter with flip flops
Simple table provides what's needed:

$$\begin{array}{c|c}
\text{Beginning State} &amp; \text{Ending State}\\\\
  {\begin{array}{cccc}
  Q_D &amp; Q_C &amp; Q_B &amp; Q_A\\\\
  0&amp;0&amp;0&amp;0\\
  0&amp;0&amp;1&amp;1\\
  0&amp;1&amp;1&amp;0\\
  1&amp;0&amp;0&amp;1\\
  1&amp;1&amp;0&amp;0\\
  1&amp;1&amp;1&amp;1\\
  0&amp;0&amp;1&amp;0\\
  0&amp;1&amp;0&amp;1\\
  1&amp;0&amp;0&amp;0\\
  1&amp;0&amp;1&amp;1\\\\
  0&amp;0&amp;0&amp;1\\
  0&amp;1&amp;0&amp;0\\
  0&amp;1&amp;1&amp;1\\
  1&amp;0&amp;1&amp;0\\
  1&amp;1&amp;0&amp;1\\
  1&amp;1&amp;1&amp;0\\
  \end{array}} &amp;
  {\begin{array}{cccc}
  D_D &amp; D_C &amp; D_B &amp; D_A\\\\
  0&amp;0&amp;1&amp;1\\
  0&amp;1&amp;1&amp;0\\
  1&amp;0&amp;0&amp;1\\
  1&amp;1&amp;0&amp;0\\
  1&amp;1&amp;1&amp;1\\
  0&amp;0&amp;1&amp;0\\
  0&amp;1&amp;0&amp;1\\
  1&amp;0&amp;0&amp;0\\
  1&amp;0&amp;1&amp;1\\
  0&amp;0&amp;0&amp;0\\\\
  x&amp;x&amp;x&amp;x\\
  x&amp;x&amp;x&amp;x\\
  x&amp;x&amp;x&amp;x\\
  x&amp;x&amp;x&amp;x\\
  x&amp;x&amp;x&amp;x\\
  x&amp;x&amp;x&amp;x\\
  \end{array}}
\end{array}$$

Now the four K-map tables.

$$\begin{array}{rl}
\begin{smallmatrix}\begin{array}{r|cccc}
D_D&amp;\overline{Q_B}\:\overline{Q_A}&amp;\overline{Q_B}\: Q_A&amp;Q_B \:Q_A&amp;Q_B \:\overline{Q_A}\\
\hline
\overline{Q_D}\:\overline{Q_C}&amp;0&amp;x&amp;0&amp;0\\
\overline{Q_D}\:Q_C&amp;x&amp;1&amp;x&amp;1\\
Q_D\: Q_C&amp;1&amp;x&amp;0&amp;x\\
Q_D\:\overline{Q_C}&amp;1&amp;1&amp;0&amp;x
\end{array}\end{smallmatrix}
&amp;
\begin{smallmatrix}\begin{array}{r|cccc}
D_C&amp;\overline{Q_B}\:\overline{Q_A}&amp;\overline{Q_B}\: Q_A&amp;Q_B \:Q_A&amp;Q_B \:\overline{Q_A}\\
\hline
\overline{Q_D}\:\overline{Q_C}&amp;0&amp;x&amp;1&amp;1\\
\overline{Q_D}\:Q_C&amp;x&amp;0&amp;x&amp;0\\
Q_D\: Q_C&amp;1&amp;x&amp;0&amp;x\\
Q_D\:\overline{Q_C}&amp;0&amp;1&amp;0&amp;x
\end{array}\end{smallmatrix}\\\\
\begin{smallmatrix}\begin{array}{r|cccc}
D_B&amp;\overline{Q_B}\:\overline{Q_A}&amp;\overline{Q_B}\: Q_A&amp;Q_B \:Q_A&amp;Q_B \:\overline{Q_A}\\
\hline
\overline{Q_D}\:\overline{Q_C}&amp;1&amp;x&amp;1&amp;0\\
\overline{Q_D}\:Q_C&amp;x&amp;0&amp;x&amp;0\\
Q_D\: Q_C&amp;1&amp;x&amp;1&amp;x\\
Q_D\:\overline{Q_C}&amp;1&amp;0&amp;0&amp;x
\end{array}\end{smallmatrix}
&amp;
\begin{smallmatrix}\begin{array}{r|cccc}
D_A&amp;\overline{Q_B}\:\overline{Q_A}&amp;\overline{Q_B}\: Q_A&amp;Q_B \:Q_A&amp;Q_B \:\overline{Q_A}\\
\hline
\overline{Q_D}\:\overline{Q_C}&amp;1&amp;x&amp;0&amp;1\\
\overline{Q_D}\:Q_C&amp;x&amp;0&amp;x&amp;1\\
Q_D\: Q_C&amp;1&amp;x&amp;0&amp;x\\
Q_D\:\overline{Q_C}&amp;1&amp;0&amp;0&amp;x
\end{array}\end{smallmatrix}
\end{array}$$

Let's start with \$D_A\$ and just follow along to see how I changed the \$x\$ values. Here we get: \$D_A=\overline{Q_A}\$:

$$\begin{array}{rl}
\begin{smallmatrix}\begin{array}{r|cccc}
D_A&amp;\overline{Q_B}\:\overline{Q_A}&amp;\overline{Q_B}\: Q_A&amp;Q_B \:Q_A&amp;Q_B \:\overline{Q_A}\\
\hline
\overline{Q_D}\:\overline{Q_C}&amp;1&amp;0&amp;0&amp;1\\
\overline{Q_D}\:Q_C&amp;1&amp;0&amp;0&amp;1\\
Q_D\: Q_C&amp;1&amp;0&amp;0&amp;1\\
Q_D\:\overline{Q_C}&amp;1&amp;0&amp;0&amp;1
\end{array}\end{smallmatrix}
\end{array}$$

Next is \$D_B\$. Again, spot my changes to \$x\$. See that: \$D_B=\overline{Q_A}\:\overline{Q_B}+Q_C\: Q_D+Q_A\: Q_B\:\overline{Q_D}\$:

$$\begin{array}{rl}
\begin{smallmatrix}\begin{array}{r|cccc}
D_B&amp;\overline{Q_B}\:\overline{Q_A}&amp;\overline{Q_B}\: Q_A&amp;Q_B \:Q_A&amp;Q_B \:\overline{Q_A}\\
\hline
\overline{Q_D}\:\overline{Q_C}&amp;1&amp;0&amp;1&amp;0\\
\overline{Q_D}\:Q_C&amp;1&amp;0&amp;1&amp;0\\
Q_D\: Q_C&amp;1&amp;1&amp;1&amp;1\\
Q_D\:\overline{Q_C}&amp;1&amp;0&amp;0&amp;0
\end{array}\end{smallmatrix}
\end{array}$$

Now for \$D_C\$. Spot changes and see: \$D_C=\overline{Q_A}\:\overline{Q_B}\:Q_C+Q_A\:\overline{Q_B}\:\overline{Q_C}+Q_B\:\overline{Q_C}\:\overline{Q_D}\$:

$$\begin{array}{rl}
\begin{smallmatrix}\begin{array}{r|cccc}
D_C&amp;\overline{Q_B}\:\overline{Q_A}&amp;\overline{Q_B}\: Q_A&amp;Q_B \:Q_A&amp;Q_B \:\overline{Q_A}\\
\hline
\overline{Q_D}\:\overline{Q_C}&amp;0&amp;1&amp;1&amp;1\\
\overline{Q_D}\:Q_C&amp;1&amp;0&amp;0&amp;0\\
Q_D\: Q_C&amp;1&amp;0&amp;0&amp;0\\
Q_D\:\overline{Q_C}&amp;0&amp;1&amp;0&amp;0
\end{array}\end{smallmatrix}
\end{array}$$

And \$D_D\$: \$D_D=\overline{Q_B}\:Q_D+ Q_C\:\overline{Q_D}\$:

$$\begin{array}{rl}
\begin{smallmatrix}\begin{array}{r|cccc}
D_D&amp;\overline{Q_B}\:\overline{Q_A}&amp;\overline{Q_B}\: Q_A&amp;Q_B \:Q_A&amp;Q_B \:\overline{Q_A}\\
\hline
\overline{Q_D}\:\overline{Q_C}&amp;0&amp;0&amp;0&amp;0\\
\overline{Q_D}\:Q_C&amp;1&amp;1&amp;1&amp;1\\
Q_D\: Q_C&amp;1&amp;1&amp;0&amp;0\\
Q_D\:\overline{Q_C}&amp;1&amp;1&amp;0&amp;0
\end{array}\end{smallmatrix}
\end{array}$$

So the equation summary from the above work is:

$$\begin{align*}
D_A&amp;=\overline{Q_A}\\
D_B&amp;=\overline{Q_A}\:\overline{Q_B}+Q_C\: Q_D+Q_A\: Q_B\:\overline{Q_D}\\
D_C&amp;=\overline{Q_A}\:\overline{Q_B}\:Q_C+Q_A\:\overline{Q_B}\:\overline{Q_C}+Q_B\:\overline{Q_C}\:\overline{Q_D}\\
D_D&amp;=\overline{Q_B}\:Q_D+ Q_C\:\overline{Q_D}
\end{align*}$$

Just note that the above isn't the only possible arrangements. By choosing differently for the \$x\$ values, you might come up with different (but equivalent) equations. If you put in a little time formulating them in several ways, you might find a better arrangement for the final circuit (fewer gates.) But I've only so much time to apply and I'm stopping it here.

Clearly, \$D_A\$ is free and \$D_D\$ is just a mux (if allowed.) The other two will involve a little more logic. But if you wire it up, it should work.

Here's how I implemented it in Neemann's Digital:



Worked exactly as expected.

The one thing that is missing is the reset to a specific point. I'll leave that to you.

The async reset to 9 was simple to add. So here it is despite "leaving it to you":