The question as posed (in the book) is a bit weird, mainly because it does not state that $F$ is required to be length preserving, however for the CBC-MAC construction to make sense it clearly has to be.
But ignoring this fact for a moment, one of your observations was indeed crucial.
A MAC does in general not hide its input message.
As you point out, if $F'$ is a secure MAC, then the MAC $F$ defined as $F_k(m) := m\Vert F'_k(m)$ is also secure.
However, if the function $F$ allows recovering its input from its output, then in the CBC-MAC construction we learn one of the intermediate values.
And that is a problem.
Let's look at the case of $\ell=2$, i.e., messages have exactly two blocks, i.e., $m=m_0\Vert m_1$.
We attack the CBC-MAC by first choosing a message $m=0^n\Vert 0^n$ and querying it to the MAC oracle.
The CBC-MAC for our query will be computed as follows:
\begin{align*}
t_1 :=& F_k(0^n)\\
t_2 :=& F_k(0^n\oplus t_1) = F_k(t_1)
\end{align*}
and $t_2$ is output as the tag.
Now, if $t_2$ allows us to recover the input of $F$, then this means we learn $t_1$.
We can now output the message $m^* = t_1\Vert t_2$ and the tag $t^*=t_1$ as our forgery.
We can verify that this is indeed a valid forgery by recomputing $t^* = t^*_2$:
\begin{align*}
t^*_1 :=& F_k(m^*_1) = F_k(t_1) = t_2\\
t^*_2 :=& F_k(m^*_2\oplus t^*_1) = F_k(t_2\oplus t_2) = F_k(0) = t_1
\end{align*}
And, given that $F$ is a secure MAC and its output therefore necessarily unpredictable means that the probability of $m^*=m$ is negligible, and therefore our attacker is successful with all but negligible probability. (This can be generalized for an arbitrary message $m$, but I'll leave this as an exercise to the reader.)
The issue that remains is: How can we construct a length-preserving MAC such that we can reconstruct $t_1$ from $t_2$?
The construction $F_k(m) := m\Vert F'_k(m)$ clearly does not work, since it is not length preserving, but we can do something similar.
Let $F'_k : \{0,1\}^n \to \{0,1\}^{n/2}$ be a secure MAC.
Then we define $F$ as $F_k(x\Vert y) = y\Vert F'_k(x\Vert y)$ for $|x|=|y|=n/2$.
I leave proving that this remains a secure MAC as an exercise to the reader.
It remains to show how this $F$ allows us to reconstruct $t_1$ given $t_2$ in the above attack.
For this, observe the values of $t_1,t_2$ in this instantiation.
\begin{align*}
t_1 :=& F_k(0^n) = 0^{n/2}\Vert F'_k(0^n)\\
t_2 :=& F_k(t_1) = F'_k(0^n) \Vert F'_k(t_1)
\end{align*}
I.e., given $t_2 := a\Vert b$, we can directly see that $t_1 := 0^{n/2}\Vert a$.
Therefore the attack sketched above works with this instantiation.
