In case anyone's still reading this, here's an elaboration of Martin Argerami's answer. In short, your formula isn't in any way a formula for actually computing anything, but rather an encapsulation of the proof of Riesz's theorem. Fleshing things out, if $f \in H^\ast$ is non-zero:
- On entirely abstract grounds, $\mathbb{C} = \operatorname{Ran}(f) \cong H/\operatorname{Ker}(f) \cong \operatorname{Ker}(f)^\perp$, so that $\operatorname{Ker}(f)^\perp$ is $1$-dimensional, and hence $\operatorname{Ker}(f)^\perp = \mathbb{C} z_0$ for any non-zero $z_0 \in \operatorname{Ker}(f)^\perp$, so fix some (viz, any) such $z_0$ as a basis for $\operatorname{Ker}(f)^\perp$. Note that this is all that one knows about what $z_0$ is.
- Writing $H = \operatorname{Ker}(f) \oplus \operatorname{Ker}(f)^\perp$, we see that $f$ is entirely determined by its restriction to $\operatorname{Ker}(f)^\perp$. However, $\operatorname{Ker}(f)^\perp$ is $1$-dimensional, so by finite-dimensional linear algebra, its dual $\left(\operatorname{Ker}(f)^\perp\right)^\ast$ is also $1$-dimensional, and thus spanned by any non-zero functional, e.g., $f_0 : x \mapsto \left\langle x,z_0 \right\rangle$. Thus, there exists some constant $\alpha \in \mathbb{C}$ such that $f|_{\operatorname{Ker}(f)^\perp} = \alpha f_0$. Plugging in our distinguished (but not explicitly known!) vector $z_0$, we find that $f(z_0) = \alpha f_0(z_0) = \alpha \left\langle z_0,z_0\right\rangle$, and hence that $$\alpha = \frac{f(z_0)}{\left\langle z_0,z_0 \right\rangle}.$$
- Putting everything together, one has that $f(x) = \left\langle x, \overline{\alpha}z_0 \right\rangle$ for $x \in \operatorname{Ker}(f)^\perp$, and hence for all $x \in H$, yielding your formula.
So, to cut a long story short, the formula you gave for $z$ depends entirely on a vector $z_0$ whose existence, guaranteed on purely abstract grounds, is the only thing we know about it. As a result, unless by some miracle you should happen to know an explicit non-zero vector in $\operatorname{Ker}(f)^\perp$, in which case you can take that as your $z_0$, the formula for $z$ is of no real computational use to you at all. In particular, it cannot possibly help you get an explicit formula for the adjoint of an operator.
In this specific case you can find the explicit formula for $T$: $u(t)=e^{At}u_{0}$. Here $e^{At}$ is defined in terms of usual Taylor expansion of the exponential function, evaluated at the matrix $At$. By Picard-Lindelof theorem the solution $u(t)$ must be uniquely given, and the particular choice $u(t)=e^{At}u_{0}$ satisfies the equation so it must be the unique solution. (Or, you can directly show it must be the solution by verifying that $e^{-At}u(t)$ must have zero derivative, so must be a constant.)
Therefore, in your equation
$$
\left\langle T^{*}v,u_{0} \right\rangle_{2}
= \int_{0}^{1}v(t)\cdot u(t)\,dt,
$$
the right-hand side is equal to
$$
\int_{0}^{1} v(t)\cdot e^{At}u_{0}\,dt = \left(\int_{0}^{1}e^{A^{*}t}v(t)\,dt\right)\cdot u_{0}
$$
(here we used the identity $(e^{At})^{*}=e^{A^{*}t}$), thus you conclude
$$
T^{*}v = \int_{0}^{1}e^{A^{*}t}v(t)\,dt.
$$
Probably it is also possible to obtain a similar formula when the ODE is of more general form. Not so sure what precisely such a statement would be.
Edit
Regarding your comment. Consider the ODE
\begin{align*}
w'(t) &= -A^{*}w(t) - v(t), \\
w(1) &= 0.
\end{align*}
Then with some similar computations, you can see that the solution is given as
$$
w(t) = e^{-A^{*}t}\int_{t}^{1}e^{A^{*}\tau}v(\tau)\,d\tau.
$$
Then $T^{*}v$ is precisely $w(0)$. In other words, you are considering the linear ODE with the coefficient matrix $A^{*}$ instead of $A$, while the input $v$ acts as the "forcing term", and you are solving the ODE "backward", so you are fixing the "final value" instead of the "initial value". The minus signs in the ODE also accounts for the direction of time being backward. Then the output of $T^{*}$ is the initial value of the ODE.
(The "general case" I mentioned in the end of the answer before the edit probably would be something similar.)
Now $T^{*}$ being given in terms of the solution $w$ of the ODE above can be derived directly as follows, which is probably what you want. Since
$$
v(t) = -A^{*}w(t) - w'(t),
$$
we can write
\begin{align*}
\left\langle T^{*}v,u_{0} \right\rangle_{2}
&= -\int_{0}^{1}A^{*}w(t)\cdot u(t) + w'(t)\cdot u(t)\,dt \\
&= -\int_{0}^{1}w(t)\cdot Au(t) + w'(t)\cdot u(t)\,dt \\
&= -\int_{0}^{1}(w(t)\cdot u(t))'dt \\
&= -(w(1)\cdot u(1) - w(0)\cdot u(0)) \\
&= w(0)\cdot u_{0},
\end{align*}
thus we get $T^{*}v = w(0)$.
Best Answer
Your proposed method is perfectly fine. You need change the order of the integrals: $$ \langle Su, v \rangle = \int_0^1 dx \int_0^xdy \ y u(y) v(x) = \int_0^1 dy \int_y^1 dx \ y u(y) v(x) = \langle u, S^\star v\rangle. $$ So $$ (S^\star v)(y) = y\int_y^1 dx \ v(x). $$