Recall that the set of functions $\{e^{2\pi int} : n \in \mathbb{Z}\}$ is an orthonormal basis for $L^2[0,1]$. Denote $e_n(t) = e^{2\pi int}$ and calculate:
$$(Ae_n)(t) = \begin{cases}
e_n(2t), & t\in[0, 1/2]\\\
e_{n}(2t-1), & t\in(1/2, 1]
\end{cases} = \begin{cases}
e^{4\pi int}, & t\in[0, 1/2]\\\
e^{4\pi int - 2\pi in}, & t\in(1/2, 1]
\end{cases} = e^{4\pi int} = e_{2n}(t)$$
Therefore $A$ acts on the othornormal basis as $Ae_n = e_{2n}$ for all $n \in \mathbb{Z}$.
Note that $A$ is isometric:
$$\|Af\|^2 = \left\|\sum_{n\in\mathbb{Z}} \langle f, e_n\rangle e_{2n}\right\|^2 = \sum_{n\in\mathbb{Z}}\left| \langle f, e_n\rangle \right|^2 = \|f\|^2$$
so $\|A\| = 1$ and hence $\sigma(A) \subseteq \overline{B}(0,1)$.
You have already established that $0$ is not an eigenvalue so let $\lambda \in \sigma_p(A)$, $\lambda \ne 0$ and let $f \ne 0$ be an eigenvector. We have
$$0 = (A - \lambda I)f = \sum_{n\in\mathbb{Z}}\langle f, e_n\rangle e_{2n} - \lambda \sum_{n\in\mathbb{Z}}\langle f, e_n\rangle e_{n} = \sum_{n\in\mathbb{Z}}\left(\langle f, e_n\rangle - \lambda\langle f, e_{2n}
\rangle\right)e_{2n} + \lambda\sum_{\substack{n\in\mathbb{Z}\\n \text{ odd}}}\langle f, e_n\rangle e_n$$
Hence $\langle f, e_n\rangle = 0$ for all odd $n\in\mathbb{Z}$ and $\langle f, e_n\rangle = \lambda\langle f, e_{2n}\rangle$ for all $n\in\mathbb{Z}$. Consequently $\langle f, e_n\rangle = 0$ when $n$ is not a power of $2$, and $\langle f, e_{\pm e_{2^k}}\rangle = \frac1{\lambda^k}\langle f, e_{\pm 1}\rangle$ for all $k \in \mathbb{N}$.
We have
$$\|f\|^2 = \sum_{n\in\mathbb{Z}}\left| \langle f, e_n\rangle \right|^2 = \left(\left|\langle f, e_1\right\rangle|^2 + \left|\langle f, e_{-1}\right\rangle|^2\right)\sum_{n\in\mathbb{N}}\frac1{|\lambda|^{2n}} + \left|\langle f, e_0\right\rangle|^2$$
The sum $\frac1{|\lambda|^{2n}}$ cannot converge because $|\lambda| \le 1$ so $\left\langle f, e_1\right\rangle = \left\langle f, e_{-1}\right\rangle = 0$ and hence $\langle f, e_n\rangle = 0$ for all $n \in \mathbb{Z} \setminus \{0\}$. Therefore $f = \langle f, e_0\rangle e_0$ so $f$ is a constant function. $Af = \lambda f$ now gives $\lambda = 1$.
Therefore $\sigma_p(A) = \{1\}$.
Now let $\lambda \in S(0,1)\setminus\{1\}$. Assume $f \perp \operatorname{Im}(A - \lambda I)$. We have $f \perp (A - \lambda I) e_n = e_{2n} - \lambda e_{n}$ which implies $\langle f, e_{2n}\rangle = \lambda\langle f, e_n\rangle$ for all $n \in \mathbb{Z}$.
This yields $\langle f, e_{\pm 2^k}\rangle = \lambda^k \langle f, e_{\pm 1}\rangle$ for all $k \in \mathbb{N}$. Left side converges to $0$ when $n\to\infty$ so $\lambda^k \langle f, e_{\pm 1}\rangle \xrightarrow{k\to\infty} 0$ also. If $|\lambda| = 1$, we conclude $\langle f, e_{\pm 1}\rangle = 0, \forall k \in \mathbb{N}$ and also $(1-\lambda)\langle f, e_0\rangle = 0$ so $\langle f, e_0\rangle = 0$. Therefore $\langle f, e_n\rangle = 0$ for all $n\in\mathbb{Z}$ so $f = 0$.
We conclude that $\overline{\operatorname{Im}(A - \lambda I)} = L^{2}[0,1]$ when $|\lambda| = 1$.
On the other hand, if $|\lambda| < 1$ then check that $f = \sum_{n=1}^\infty \lambda^k e_{2^k} \perp \operatorname{Im}(A - \lambda I)$ so $\overline{\operatorname{Im}(A - \lambda I)} \ne L^{2}[0,1]$. We conclude that $\sigma_r(A) = B(0,1)$.
The spectrum is a closed set so we conclude $\sigma(A) = \overline{B}(0,1)$. Finally, $\sigma_c(A) = S(0,1) \setminus \{1\}$.
To conclude:
$$\sigma(A) = \overline{B}(0,1)$$
$$\sigma_p(A) = \{1\}$$
$$\sigma_r(A) = B(0,1)$$
$$\sigma_c(A) = S(0,1)\setminus \{1\}$$
