Computing the dual change of coordinate matrix $[T^t]^{beta *}_{gamma *}$

Above they already showed you the reasoning, but I want to show you another way of doing it. I will first observe one thing in "abstract".

If $V$ is a finite-dimensional vector space, $alpha = {v_1,dots,v_n}$ is a basis for $V$, and $alpha^* = {phi_1,dots,phi_n}$ the corresponding dual basis, then any $f in V^*$ can be written as $f = f(v_1)phi_1 + cdots + f(v_n) phi_n$.

This is easy to see, for if $v in V$, then $v = phi_1(v)v_1 + cdots + phi_n(v)v_n$, and then $$f(v) = phi_1(v)f(v_1) + cdots + phi_n(v)f(v_n) = big( f(v_1)phi_1 + cdots + f(v_n) phi_n big)(v).$$ So, in this concrete example we have to write the linear functionals $T^t(g_1)$ and $T^t(g_2)$ as a linear combination of $f_1$ and $f_2$, and because ${f_1,f_2}$ is the dual basis of ${1,x}$ we have: begin{align} T^t(g_1) &= T^t(g_1)(1)f_1 + T^t(g_1)(x)f_2 \ &= g_1(T(1))f_1 + g_1(T(x))f_2 \ &= g_1(1,1)f_1 + g_1(0,2)f_2 \ &= 1f_1 + 0f_2 end{align} and similarly, $T^t(g_2) = g_2(1,1)f_1 + g_2(0,2)f_2 = 1f_1 + 2f_2$.

I believe that your main issue is that you are used to think of bases in an abstract fashion. That is, if $beta:={x_1, ldots, x_n}$ is a basis for a vector space $X$ then the dual basis $beta^*={f_1, ldots, f_n}$ are linear functionals such that $f_{i}(x_j)=delta_{i,j}$. However, for this question you have some concrete vector spaces and some well known bases for each of them.

First of all since $beta$ is the standard ordered bases for $P_1(Bbb{R})$ we actually have $beta={1, x}$. Thus, the dual basis is $beta^*={f_1, f_2}$, where $f_1, f_2 : P_1(Bbb{R}) to Bbb{R}$ are such $f_1(1)=1$, $f_1(x)=0$, $f_2(1)=0$ and $f_2(x)=1$ (think of $1$ as $x_1$ and $x$ as $x_2$ in the abstract fashion above). Hopefully this answers one of your questions.

Similarly, $gamma={(1,0), (0,1)}$ is the standard basis for $Bbb{R}^r$ and therefore the dual basis is $gamma^*:={ g_1 ,g_2}$ where $g_1, g_2: Bbb{R}^2 to Bbb{R}$ are such that $g_1(1,0)=1$, $g_1(0,1)=0$, $g_2(1,0)=0$ and $g_2(0,1)=1$ (think of $(1,0)$ as $x_1$ and $(0,1)$ as $x_2$ in the abstract fashion above). Therefore, since $g_1$ is linear $$ g_1(1,1)=g_1( (1,0)+(0,1) ) = g_1(1,0)+g_1(0,1)=1+0=1 $$ This should answer what $g(1,1)$ is and why is it equal to $1$.

Finally your main goal is to find the entries $a,b,c$ and $d$ for the matrix of the linear transformation $T^t$ with respect to the bases $gamma^*$ and $beta ^*$. To do this you have to use that there are two ways to compute $T^t(g_1)(1)$, namely

1. Using the matrix: $T^t(g_1)(1)=(af_1+cf_2)(1)=a+0=a$
2. By definition of $T^t$: $T^t(g_1)(1)=g_1(T(1))= g_1(1,1) = 1$

This gives you the value of $a$. Analogously there are two ways to compute $T^t(g_1)(x)$, namely

1. Using the matrix: $T^t(g_1)(x)=(af_1+cf_2)(x)=c$ (because $f_1(x)=0$ and $f_2(x)=1$)
2. By definition of $T^t$: $T^t(g_1)(x)=g_1(T(x))= g_1(0,2) = 2g_1(0,1)= 0$

This now gives the value of $c$. Similarly, when computing both $T^t(g_2)(1)$ and $T^t(g_2)(x)$ using matrix way and the definition way you should be able to find the values for $b$ and $d$.

Do you think you can take it from here now?

I hope this is helpful.