The problem with your argument is that a sequence with length $n - 2$ that does not contain two consecutive zeros but ends with a zero can be extended to one that does contain two consecutive zeros by appending a single zero, which places it among the $a_{n - 1}$ admissible strings of length $n - 1$.
Let's examine admissible strings of length $4$. They are the eight strings
$$0000, 0001, 0010, 0100, 1000, 0011, 1001, 1100$$
Notice that since the admissible strings of length $3$ are
$$000, 001, 100$$
you counted the six strings
$$0000, 0001, 0010, 0011, 1000, 1001$$
in your $2a_3$ admissible strings of length $4$ that can be formed by appending a $0$ or $1$ to the end of an admissible string of length $3$.
Since the inadmissible strings of length $2$ are
$$01, 10, 11$$
you counted the three strings
$$0100, 1000, 1100$$
among the $2^{n - 2} - a_{n - 2}$ among the admissible strings of length $4$ that can be formed by appending $00$ to an inadmissible string of length $2$.
Notice that you have counted the string $1000$ twice since $100$ is an admissible string of length $3$ and $10$ is an inadmissible string of length $2$. The problem, as stated above, is that $10$ is an inadmissible string of length $2$ that can be extended to an admissible string of length $3$ by appending a $0$ to the end of an inadmissible string of length $2$ that ends in a zero.
To find these values by hand you can use the concept of dynamic programming.
Lets first denote the following expressions:
$S_{n, 0}$ - number of valid sequences of length $n$ starting with $0$
$S_{n, 1}$ - number of valid sequences of length $n$ starting with $1$
$S_{n}$ - number of valid sequences of length $n$
Obviously $S_n = S_{n, 0} + S_{n, 1}$
Let's observe that any such sequence must have $0$ as the second digit.
If sequence of length $n$ begins with $1$ then the remaining part of the sequence must be valid sequence of length $n - 1$ and must begin with $0$. So:
$$S_{n, 1} = S_{n-1, 0}$$
Furthermore if sequence of length $n$ begins with $0$ then we can have two cases.
Case $1$: third digit of the sequence is equal to $0$
We can easily denote that the number of such sequences is equal to $S_{n-1, 0}$, becuase apart from starting $0$ the remainig part must form valid sequence of lentgh $n-1$ that starts with $0$.
Case $2$: third digit of the sequence is equal to $1$
We can observe that, this $1$ already has adjacent $0$ so we can say that remaining part of the sequence (apart from the first three digits) must form valid seqeunce of length $n-3$ and there are $S_{n-3}$ such sequences.
So we can denote:
$$S_{n, 0} = S_{n-1, 0} + S_{n-3}$$
If we now take everything together we will get:
$$S_n = S_{n, 0} + S_{n, 1} = S_{n-1, 0} + S_{n-1, 0} + S_{n-3} = 2S_{n-1, 0} + S_{n-3}$$
To efficiently calculate those values we can store these results in a table:
$$
\begin{array}{|c|c|c|c|c|}
\hline
n & 3 & 4 & 5 & 6 & 7 & \cdots & \cdots & \cdots & \cdots \\
\hline
S_{n, 0} & 2 & 2 & 3 & 6 & 10 & \color{red}{15}& \\
\hline
S_{n-2} & ? & ? & 3 & 4 & 5 & 9 & 16 & \color{green}{25}&\\
\hline
\end{array}
$$
To fill up the table you have to first find the value of first row by adding values in previous column (here you get $10 + 5 = 15$) and then you can find the value in the second row by adding to previously achieved result value from the second last cell in the first row (here $15 + 10 = 25$). Notice that there is a shift applied for the values of $S_n$ beacuse now you can fill the table in more convenient way.
Best Answer
Here is a full solution. First of all, note that your sequence of strings is very similar to the Thue-Morse sequence. The only difference is that, if $n$ is even, a string in your sequence is inverted ($0$s become $1$s, and vice versa). Similar to the Thue-Morse sequence, your sequence can be constructed in a similar way. To get a $n+1$-st string one should take an $n$-th string and add an inverted $n$-th string to the beginning of it.
We will prove that by induction. The first string transformation $1\to01$ respects this rule. Consider a string $\overline ss$, which was transformed from $s$. (We denote a string where $1$s are substituted by $0$s and vice versa by $\overline s$.) What will happen next? The right half of $\overline ss$, which is $s$, will becomes $\overline ss$, as we already know. The left half $\overline s$ is inverted $s$, and since the rules $0\to 10$ and $1\to 01$ are inversions of each other, $\overline s$ will become $s\overline s$. Hence, $\overline ss$ will become $s\overline s\overline ss$, just as we wanted to prove.
This means that all substrings $00$ are preserved when go to the next string, also all $11$s become $00$s in the first half of the new string, too. And there can be one more $00$ in the place of concatenation. It is easy to see, that for even $n$ a string begins and ends in $1$. And for odd $n$ a string begins by $0$ and ends by $1$. So there appears one more $00$ when we go from an odd string to the next one.
Let us denote the number of $00$s in the $n$-th string by $a_n$, the number of $11$s in the $n$-th string by $b_n$ ($n$ is the number of a string as in the OP’s table, or the number of transformations). Then the following is true (for $n\ge0$):
$$\begin{align} a_{n+1}&=a_n+b_n+e_n\\ b_{n+1}&=a_n+b_n,\end{align}$$
where $e_n$ is $1$ if $n$ is odd, and $0$ otherwise. In other words, $e_n=\frac12+\frac12(-1)^{n-1}$. One can see that the following is true for $a_n$, $b_n$ (for $n\gt0$):
$$\begin{align} a_{n+1}&=2a_n+(-1)^{n+1}\\ b_{n+1}&=2b_n+1-e_n.\end{align}$$
Indeed, we have $$a_{n+1}=a_n+b_n+e_n=a_n+(a_n-e_{n-1})+e_n=$$ $$=2a_n-\frac12-\frac12(-1)^{n-2}+\frac12+\frac12(-1)^{n-1}=2a_n+(-1)^{n-1}.$$
The second equation is not really needed for our goal. But we can see, that $$b_{n+1}=a_n+b_n=(b_n+e_{n-1})+b_n=2b_n+1-e_n.$$
Now we just have to “solve” the recursion for $a_n$. If the equation were homogeneous, then the solution would be $a_n=k\cdot2^n$. The particular solution of the non-homogeneous equation has a form $l\cdot(-1)^n$. Putting this into the equation we get $$l\cdot(-1)^{n+1}=2l\cdot(-1)^n+(-1)^{n+1}.$$
Dividing by $(-1)^n$ we get $l\cdot(-1)=2l+(-1)$, so $l=\frac13$. Hence $a_n=k\cdot 2^n+\frac{(-1)^n}3$ for some $k$. Knowing that $a_1=0$ we find: $$0=k\cdot2-\frac13.$$
So, $k=\frac16$. And finally, for $n\gt0$: $$\boxed{a_n=\frac{2^{n-1}+(-1)^n}3.}$$
Out of curiosity, we can find $b_n$:
$$b_n=a_{n}-e_{n-1}=\frac{2^{n-1}+(-1)^n}3-\frac12-\frac12(-1)^{n}=$$
$$=\frac{2^{n-1}-\frac32-\frac12(-1)^n}3.$$