Although the answer of @SRKX is right on spot, I was already writing a solution along the lines of how you had specifically approached the problem. I think it might still be useful to you, so here it goes
The price of the chooser option, as seen of today $t=0$ is by definition
\begin{align}
V_0 &= \underbrace{e^{-r T_2}}_{\text{Payoff dicount factor}} \underbrace{\mathbb{E}\left[\ \ \underbrace{\max\left( \mathbb{E}[(S_{T_2}-K)^+ \vert \mathcal{F}_{T_1}], \mathbb{E}[(K-S_{T_2})^+ \vert \mathcal{F}_{T_1} ] \right)}_{\text{Expected payoff at $T_2$ as seen of $T_1$}} \ \ \vert \mathcal{F}_0 \ \ \right]}_{\text{Expected payoff at $T_2$ as seen of $t=0$}} \\
&= e^{-r T_2} \mathbb{E}_0\left[ \max\left( \mathbb{E}_{T_1}[(S_{T_2}-K)^+], \mathbb{E}_{T_1}[(K-S_{T_2})^+] \right) \right]
\end{align}
If you're not familiar with the notation $\mathcal{F}_t$ used for filtrations, you can interpret it as "all the information we know at time $t$". The notation $\mathbb{E}_t[.]$ simply figures that the expectation is taken conditionally on the knowledge of $\mathcal{F}_t$. Naturally all of these expectations are taken under the risk-neutral measure $\mathbb{Q}$.
By definition, we also have that the price of European call/put options is given by
$$ C(T_1,S_{T_1};K,(T_2-T_1)) = e^{-r(T_2-T_1)} \mathbb{E}_{T_1}[(S_{T_2}-K)^+] \tag{def 1} := C_{12} $$
$$ P(T_1,S_{T_1};K,(T_2-T_1)) = e^{-r(T_2-T_1)}\mathbb{E}_{T_1}[(K-S_{T_2})^+] \tag{def 2} := P_{12} $$
where $C(t,S_t;K,\tau)$ (resp. $P(t,S_t;K,\tau)$) denotes the price of a European call (resp. put) option as seen of time $t$, given the spot value $S_t$, the strike price $K$ and the time to expiry $\tau$.
Therefore,
$$ V_0 = e^{-r T_2} \mathbb{E}_0\left[ \max\left( \frac{C_{12}}{e^{-r(T_2-T_1)}}, \frac{P_{12}}{e^{-r(T_2-T_1)}} \right) \right]$$
Yet by call-put parity:
$$ C_{12} - P_{12} = e^{-r(T_2-T_1)}( S_1 e^{(r-q)(T_2-T_1)} - K ) $$
so that we can further write (similarly to what you did)
\begin{align}
V_0 &= e^{-r T_2} \mathbb{E}_0\left[ \max\left( \frac{C_{12}}{e^{-r(T_2-T_1)}}, \frac{C_{12}}{e^{-r(T_2-T_1)}} - (S_1e^{(r-q)(T_2-T_1)} - K) \right) \right] \\
&= e^{-r T_2} \mathbb{E}_0\left[ \left( \frac{C_{12}}{e^{-r(T_2-T_1)}} + \max\left( 0, K - S_1e^{(r-q)(T_2-T_1)} \right) \right) \right] \\
&= \mathbb{E}_0\left[ e^{-rT_1} C_{12} \right] + \mathbb{E}_0\left[ e^{-rT_2} \max\left( 0, K - S_1e^{(r-q)(T_2-T_1)} \right) \right] \tag{1}
\end{align}
Now using $(\text{def } 1)$ the first term of $(1)$ becomes:
$$ \mathbb{E}_0 \left[ e^{-rT_1} C_{12} \right] = \mathbb{E}_0 \left[ e^{-rT_2} \mathbb{E}_{T_1}[(S_{T_2}-K)^+] \right] = C(0,S_0;K,T_2)$$
by the tower property of conditional expectations.
Similarly, the second term of $(1)$ can on the other hand be expressed as:
\begin{align}
\mathbb{E}_0\left[ e^{-rT_2} \max\left( 0, K - S_1e^{(r-q)(T_2-T_1)} \right) \right]
&= \mathbb{E}_0\left[ \max\left( 0, Ke^{-rT_2} - S_1e^{-rT_1-q(T_2-T_1)} \right) \right] \\
&= e^{-q(T_2-T_1)} \mathbb{E}_0\left[ e^{-r{T_1}} \max\left( 0, Ke^{-(r-q)(T_2-T_1)} - S_1 \right) \right] \\
&= e^{-q(T_2-T_1)} P(0,S_0; Ke^{-(r-q)(T_2-T_1)}, T_1)
\end{align}
So that $(1)$ becomes
$$ V_0 = C(0,S_0;K,T_2) + \underbrace{e^{-q(T_2-T_1)}}_{= 0.9778} P(0,S_0; \underbrace{Ke^{-(r-q)(T_2-T_1)}}_{= 62.1085}, T_1) $$
hence a $T_2$ call struck at $K$ + 0.9778 units of a $T_1$ put with adjusted strike 62.1085.
