$L:L_2[0,T] \to \mathbb R^n$
$$ Lf = \int_0^T \exp(A(T-t))\pmb b f(t)\;dt $$
I'm supposed to show that $L^*=\pmb b^T \exp(A^T(T-t))$.
The inner product is over $\mathbb R^n$ since $Lf \in \mathbb R^n$.
$$ \begin{align*} \langle Lf, \pmb y \rangle &= \pmb y^T\left(\int_0^T \exp(A(T-t))\pmb bf(t)\;dt\right) \end{align*} $$
It's be easy if I could separate $f(t)$ from $\int_0^T \exp(A(T-t))$ but since these are multiplied together I can't distrubte the integral through so I don't know how to disentangle $f(t)$ from the $L$ part.
You're right that on one side of the equation of adjointness it's the inner product of $\Bbb R^n$, however the other side should utilize the inner product of $L^2([0,T])$, hence the integral would be introduced, and luckily we don't need to carry out $f(t)$ or the exponential part, but move inside the inner product with ${\bf y}$.
Specifically, we have to prove $\langle Lf,\, {\bf y}\rangle_{\Bbb R^n}\ =\ \langle f,\, L^*{\bf y}\rangle_{L^2([0,T])}$.
Using the suggested $L^*$, the right hand side becomes $$\int_0^T f(t)\cdot{\bf b}^\top\exp(A^\top(T-t)){\bf y}\, dt$$
Now observe that ${\bf b}^\top\exp(A^\top(T-t)){\bf y}$ is a scalar (for each $t\in [0,T]$), and hence it equals to its own transpose ${\bf y}^\top\exp(A(T-t)){\bf b}$.