Let X be a Banach space with a closed subspace Y, We define the dual mapping $ \pi^*:(X/Y)^* → X^*$ by $\pi^* (\beta)=\beta\circ\pi$ then $(X/Y)^*$is isometrically isomorphic to $$Y^\perp:=\{f∈ X^*|f(Y)=\{0\}\}$$
we define $T:(X/Y)^* → Y^\perp$. we must show that $T$ is a linear isomorphism such that $T$ is a isometry from $(X/Y)^*$ to $Y^\perp $.
$T=\pi^*$ (with restricted range). Note that for $ β∈(X/Y)^∗$, and $y∈Y$ we have $$π^∗β(y)=β(πy)=β(0)=0$$ Hence$ π^∗[(X/Y)^∗]⊆Y^⊥$.
Yes, you are on the right track. Actually your $T$ is an isometric isomorphism.
First let me prove that $T$ is an isometry. That is, we have to prove that $\lVert \beta\circ\pi\rVert = \lVert\beta\rVert$ for all $\beta\in (X/Y)^*$. Note that
$$\lVert\beta\rVert = \sup_{\lVert x+Y\rVert < 1}\lvert\beta(x+Y)\rvert = \sup_{\lVert x+Y\rVert < 1} \lvert\beta\circ\pi(x) \rvert = \sup_{\lVert x\rVert < 1} \lvert \beta\circ\pi(x)\rvert,$$ because $\pi(B_X[0,1)) = B_{X/Y}[0,1)$ where the balls are open balls. Notice that $\sup\limits_{\lVert x\rVert < 1} \lvert\beta\circ\pi(x)\rvert = \lVert \beta\circ\pi\rVert$. Hence it is an isometry.
Now we will prove the surjectivity of $T$. Let $f\in Y^{\perp}$. Then by using the fact that $f(Y) = 0$ and $f\in X^*$, we can easily verify that the $\phi\colon X/Y \to \mathbb{K}$ defined by $\phi(x+Y) = f(x)$ is a well defined continuous linear functional and $\phi\circ\pi = f$.