First to clarify what exactly your question is asking. Your question asks us to consider a linear map from $\mathbb{R}^2$ to $\mathbb{R}^n$, for some integer $n$. I believe the question does not ask you to consider different values of $n$, but instead concerns how a linear map from $\mathbb{R}^2$ to some vector space might behave.
Rephrasing, let $W$ be any vector space, and let $F : \mathbb{R}^2 \to W$ be a linear map. We would then like to show that either $F(A), F(B)$ are linearly independent, the image of $F$ is of dimension $1$, or $F$ is the zero map.
You are right to want to make extensive use of Theorem 3.2. It is a rather useful result, so I shall do the same. Since $F$ is a linear map from $\mathbb{R}^2$ to $\mathbb{R}^n$, $F$ is going to play the role of $L$, $\mathbb{R}^2$ will be $V$, and $\mathbb{R}^n$ will be $W$ in the theorem. The theorem then says that
$$\dim V = \dim \ker L + \dim \text{im } L.$$
In our case, this means that
\begin{align}
\dim \mathbb{R}^2 &= \dim \ker F + \dim \text{im } F\\
\implies 2 &= \dim \ker F + \dim \text{im } F \text{ (The dimension of $\mathbb{R}^2 = 2$)}.
\end{align}
So no matter how $F$ acts on $A$ and $B$, it is always true that the sum of the dimension of its kernel and the dimension of its image will be $2$. We know that the dimension of a vector space must be an integer and cannot be negative, so this restricts our list of possible choices for $\dim \ker F$. Then $\dim \ker F = 0, 1, $ or $2$ (if $2$ is the sum of two positive numbers, what other choices do we have?).
Now we just see what happens to our linear map in each case. If $\dim \ker F = 0$, then what does $F$ look like? As you mentioned before, if $\ker F = \{0\}$, then $F(A), F(B)$ are linearly independent by Theorem 3.1 (what is the dimension of the image in this case?).
If you answered the question above, then you will see that if $\dim \ker F = 1$, it follows that $\dim \text{im } F = 1$ as well (use one of your formulas from (1) to (3)).
Lastly, we have the case where $\dim \ker F = 2$. Presumably, we know that $\ker F$ is a subspace of $\mathbb{R}^2$ and since $\dim \ker F = 2$, it follows that $\ker F = \mathbb{R}^2$. What does this say about $\text{im } F$? (Take a vector in $\mathbb{R}^2$ and see what $F$ does to it, also use one of your formulas from (1) to (3) like before).
Your approach is correct!
P1) $\dim(Im \ F)=0 \implies Im(F)=\{0\}$, because the image of a linear function is a subspace and hence $0$ is in it, and can't have anything else because its dimension is zero. So $F(x)=0 \ \forall x$
P2) we have $\dim(Ker \ F)=1$, applying the theorem you get $\dim(Im \ T)=1$ and you can use the fact that two vector spaces are isomorphic (they are "the same space") if their dimension are equal, hence you can say that $Im(T)\cong \mathbb{R}$ which is a very nice way to justify that "$Im(T)$ is a straight line".
P3) can't be the case that $\dim(Ker \ T)=0$ because this would implie $Ker(T)=\{0\}$, but we know that $A\not=0$ and $A\in Ker(T)$
Your answer is good too! But it seems like it need to be more "direct" in a way... but the question isn't too direct either... I assumed that "being a straight line" is the same that "have dimension one"... but justifying that dimension one implies being isomorphic to the reals is also a good argument (because they are often called THE line).
Best Answer
If $b=0$, we don't always get $\dim f(\Bbb R^n)=n$ (especially if $m<n$).
Rather, the dimension of the image of a linear map is called its rank, which is the same as the rank of the matrix $A$: the maximum number of independent columns.
If $b\ne 0$, the image $f(\Bbb R^n)$ is in general not a linear subspace but an affine subspace, namely the column space of $A$ shifted by $b$.
As an affine subspace, its dimension is still ${\rm rank}(A)$.