When we apply a phase gate, a relative phase is added (this is the definition that I will use in this answer for the phase gate). In the Qiskit's textbook (and in the textbook by M. Nielsen and I. Chuang) $T$ is defined as a phase gate $P(\frac{\pi}{4})$:
$$
P |\psi \rangle = P (\alpha |0\rangle + \beta |1\rangle) = \alpha |0\rangle + e^{i\varphi}\beta |1\rangle \\
P(\varphi) =
\begin{pmatrix}
1&0 \\
0&e^{i \phi}
\end{pmatrix}
\qquad
T = P(\frac{\pi}{4}) =
\begin{pmatrix}
1&0 \\
0&e^{i \frac{\pi}{4}}
\end{pmatrix}
$$
where $P$ is the phase gate, $\alpha$ and $\beta$ are some initial amplitudes, $\varphi$ is the phase defined by the $P$ gate. Note that only $|1\rangle$ in the superposition state has obtained the phase. The same works for the controlled phase gate: only $|11\rangle$ obtains a phase because the control qubit should be $|1\rangle$ and the target qubit also should be $|1\rangle$:
$$CP_{2 \rightarrow 1} |+1 \rangle = CP_{2 \rightarrow 1} \frac{1}{\sqrt{2}} (|01\rangle + |11\rangle) = \\
= \frac{1}{\sqrt{2}} (|01\rangle + e^{i \varphi}|11\rangle) = \frac{1}{\sqrt{2}} (|0\rangle + e^{i \varphi}|1\rangle) \otimes |1\rangle$$
where $CP$ is the controlled phase gate, $2 \rightarrow 1$ subscript denotes that the $CP$ gate is controlled by the second qubit. More general proof for $CP_{1 \rightarrow 2} = CP_{2 \rightarrow 1}$ can be derived by using matrix representation of the $CP$ gate. This proof is similar to the proof for $CZ_{1 \rightarrow 2} = CZ_{2 \rightarrow 1}$ that can be found in this answer.
$$CP_{1 \rightarrow 2} = |0\rangle \langle 0|
\otimes I + |1\rangle \langle 1| \otimes P = \\
=
\begin{pmatrix}
1&0&0&0 \\
0&1&0&0 \\
0&0&1&0 \\
0&0&0&e^{i \varphi} \\
\end{pmatrix} = \\
=I \otimes |0\rangle \langle 0|
+ P \otimes |1\rangle \langle 1| = CP_{2 \rightarrow 1}$$
Side note about why the "symmetry" is not true for controlled $R_z$ gate in contrast to controlled $P$ gate:
If for the general case instead of $P(\varphi)$ we will use $R_z(\varphi)$ gate then we will have a different result:
$$CRZ_{1 \rightarrow 2} = |0\rangle \langle 0|
\otimes I + |1\rangle \langle 1| \otimes R_z
=
\begin{pmatrix}
1&0&0&0 \\
0&1&0&0 \\
0&0&e^{-i \frac{\varphi}{2}}&0 \\
0&0&0&e^{i \frac{\varphi}{2}} \\
\end{pmatrix}\\
CRZ_{2 \rightarrow 1} = I \otimes |0\rangle \langle 0| + R_z \otimes |1\rangle \langle 1| =
\begin{pmatrix}
1&0&0&0 \\
0&e^{-i \frac{\varphi}{2}}&0&0 \\
0&0&1&0 \\
0&0&0&e^{i \frac{\varphi}{2}} \\
\end{pmatrix}$$
where $R_z(\varphi) = \begin{pmatrix}
e^{-i \frac{\varphi}{2}}&0 \\
0&e^{i \frac{\varphi}{2}}
\end{pmatrix}$. So $CRZ_{1 \rightarrow 2} \ne CRZ_{2 \rightarrow 1}$.
This answer also might be relevant where the difference between the controlled versions of $R_z$ and $U1 = P$ is discussed.
