There are a few ways to verify it, depending on what you're more comfortable with.
The first method is to set $w = \sin z$, so that
$$
\begin{align}
\tan z &= \frac{w}{\sqrt{1-w^2}} \\
&= \sum_{k=0}^{\infty} \binom{k-\tfrac{1}{2}}{k} w^{2k+1} \\
&= \sum_{k=0}^{\infty} \binom{k-\tfrac{1}{2}}{k} (\sin z)^{2k+1},
\end{align}
$$
the first equality holding for $|z| < \pi/2$. Power series are asymptotic series, so we've got what we want.
A second method takes more advantage of power series. The first thing to note is that since both $\tan$ and $\sin$ are odd, they both only have odd powers in their expansions.
In the base case we know that $\tan z \sim \sin z$ as $z \to 0$. Since they're equal to first order only, their difference will be asymptotic to
$$
\tan z - \sin z \sim c_1 z^3 \sim c_1 (\sin z)^3
$$
for some constant $c_1$.
Now suppose that we have found some sequence of coefficients $c_0, c_1, \ldots, c_n$ with $c_0 = 1$ such that
$$
\tan z - \sum_{k=0}^{n} c_k (\sin z)^{2k+1} \sim C (\sin z)^{2N+1}.
$$
for some $C \in \mathbb{R}$ and $N \in \mathbb{N}$ with $N > n$. Then, setting $c_{n+1} = c_{n+2} = \cdots = c_{N-1} = 0$ and $c_N = C$, the new difference,
$$
\tan z - \sum_{k=0}^{N} c_k (\sin z)^{2k+1},
$$
will have a power series expansion beginning with a term of the form $Dz^{2M+1}$ for some $D \in \mathbb{R}$ and some $M \in \mathbb{N}$ with $M > N$. We then have
$$
\tan z - \sum_{k=0}^{N} c_k (\sin z)^{2k+1} \sim Dz^{2M+1} \sim D(\sin z)^{2M+1}.
$$
Proceeding by induction we construct the full asymptotic series for $\tan z$ in terms of powers of $\sin z$.
A third way to verify it involves proving that each power of $z$, $z^n$, has an asymptotic series in terms of powers of $\sin z$.
First we wish to show that $z$ has an asymptotic series in terms of powers of $\sin z$. To do this we can use the idea from the first method. Setting $w = \sin z$ we have
$$
\begin{align}
z &= \arcsin w \\
&= \sum_{k=0}^{\infty} \frac{\binom{2n}{n}}{4^n (2n+1)} w^{2n+1} \\
&= \sum_{k=0}^{\infty} \frac{\binom{2n}{n}}{4^n (2n+1)} (\sin z)^{2n+1}
\end{align}
$$
for $|z| < \pi/2$. Taking powers of this we can find coefficients $c_{n,k}$ such that
$$
\begin{align}
z^{2n+1} &= \left(\sum_{k=0}^{\infty} \frac{\binom{2k}{k}}{4^k (2k+1)} (\sin z)^{2k+1}\right)^{2n+1} \\
&= \sum_{k=n}^{\infty} c_{n,k} (\sin z)^{2k+1}
\end{align}
$$
for $|\sin z| < 1$.
From this point we will only use the fact that these series are asymptotic expansions, and we will only assume that the power series for $\tan z$ is asymptotic, not convergent. This is to emphasize that the manipulations from this point on will work the same way for other asymptotic series.
Fix $N \in \mathbb{N}$. We have
$$
\tan z = \sum_{n=0}^{N} b_n z^{2n+1} + O(z^{2N+3})
$$
and for each $n = 0, 1, \ldots, N$ we have
$$
z^{2n+1} = \sum_{k=n}^{N} c_{n,k} (\sin z)^{2k+1} + O\Bigl((\sin z)^{2N+3}\Bigr)
$$
as $z \to 0$. Combining these we obtain
$$
\tan z = \sum_{n=0}^{N} d_n (\sin z)^{2n+1} + O\Bigl((\sin z)^{2N+3}\Bigr) + O(z^{2N+3})
$$
for some constants $d_n$. Since $z^{2N+3} = O\Bigl((\sin z)^{2N+3}\Bigr)$ this is
$$
\tan z = \sum_{n=0}^{N} d_n (\sin z)^{2n+1} + O\Bigl((\sin z)^{2N+3}\Bigr)
$$
as $z \to 0$. Since $N$ was arbitrary, this is an asymptotic series for $\tan z$.
Using the same idea we could start at this series and find the original asymptotic series in terms of powers of $z$.
