To shed light on some questions:
The statement "$G=T\rtimes Q \Leftrightarrow Q \cong Aut(\mathcal{L}^n)$" is wrong.
The statement "if $Q \cong Aut(\mathcal{L}^n)$ and $T \cong \mathcal{L}^n$, then $G=T\rtimes Q$" is wrong.
The following is true:
Theorem: There is a 1-1 correspondence between symmorphic space groups of dimension $n$ and the conjugacy classes of finite subgroups of $GL_n(\mathbb{Z})$.
Explanations:
In the following I write $L$ instead of $\mathcal{L}^n$ for a lattice and $Aut_O(L) := Aut(L) \cap O_n$ instead of $Aut(\mathcal{L}^n)$, where $O_n$ is the orthogonal group and $Aut(L)$ is the group of automorphisms of $L$ in the group theoretic sense.
1) Choose $L := \mathbb{Z}^n$. Since columns of matrices in $O_n$ have Euklid norm 1, $Aut_O(L) =GL_n(\mathbb{Z})\cap O_n$ consists of the signed permutation matrices (there are $2^n \cdot n!$). For a subgroup $Q \le Aut_O(L)$ let $G_Q := \langle t_a, r_B \mid a \in L, B \in Q \rangle$, where $t_a$ is translation by $a$ and $r_B: \mathbb{R}^n \to \mathbb{R}^n, x \mapsto Bx$. Then $G_Q$ is the semi-direct product of its translational subgroup $T = \langle t_a \mid a \in L \rangle$ and of $\langle r_B \mid B \in Q \rangle \cong Q$. This shows that $(\Rightarrow)$ is false (just take any $Q \lvertneqq Aut_O(L)$).
In order to show that $(\Leftarrow)$ is also false, it suffices to find a counterexample in dimension 2 (I guess there are counterexamples in each dimension, but that would be a - nontrivial - topic for itself). $Aut_O(\mathbb{Z}^2)$ is the dihedral group $D_8$ of order 8 and the existence of such a counterexample follows from $H^2(D_8;\mathbb{Z}^2) = \mathbb{Z}/2$.
(Under the 17 space groups of dimension 2, there are exactly 2 with point group $D_8$ and by looking into the corresponding tables you should be able to figure out the non-split one to get an explicit presentation.)
2) This was just shown by the counterexample. You misread the explication in the Havana paper: The author merely shows the existence of a space group those point group is isomorphic to $Aut_O(L)$ - namely the semi-direct product (but he does not say that, if the point group is isomorphic to $Aut_O(L)$, then the space group is symmorphic!). This is to justify the subsequent definition of Bravais groups (Def. 42).
3) In the following write $L_0 := \mathbb{Z}^n$ and $T_0 = \langle t_a \mid a \in \mathbb{Z}^n \rangle$.
Let $\mathcal{C}_n$ be the set of conjugacy classes of finite subgroups
of $GL_n(\mathbb{Z})$ and let $\mathcal{I}_n$ be the set of isomorphism
classes of symmorphic space groups. Define a mapping
$$f: \mathcal{C}_n \to \mathcal{I}_n, [Q] \mapsto [G_Q]$$
where $[.]$ denotes the corresponding class on either side and $G_Q := L_0 \rtimes Q$, where $Q$ acts on $L_0$ through matrix multiplication.
Step 0: $f$ is well-defined
Let $Q,P \le GL_n(\mathbb{Z})$ be conjugated in $GL_n(\mathbb{Z})$,
i.e. there is $A \in GL_n(\mathbb{Z})$ such that $P = AQA^{-1}$. Using
$$\alpha: L_0 \to L_0, x \mapsto Ax,$$
$$\beta: Q \to P, B \mapsto ABA^{-1},$$ one immediately sees
$\beta(B)\cdot \alpha(x) = ABA^{-1}Ax = \alpha(Bx)$. Consequently
$$L_0 \rtimes Q \to L_0 \rtimes P, (x,B) \mapsto (\alpha(x),\beta(B))$$
is an isomorphism. Thus $G_Q = L_0 \rtimes Q \cong L_0 \rtimes P = G_P$.
Next I show that $G_Q$ is isomorphic to a symmorphic space group. By Maschke's theorem
$Q$ is conjugate in $GL_n(\mathbb{R})$ to a subgroup $P$ of $O_n$. So there is
$A \in GL_n(\mathbb{R})$ such that $P = AQA^{-1}$. Let $L := AL_0 \le \mathbb{R}^n$. Then the isomorphisms $\alpha: L_0 \to L$, $\beta: Q \to P$ (defined by the same formulars as above) yield $G_Q \cong L \rtimes P \cong \langle t_a \mid a \in L \rangle \rtimes \langle r_B \mid B \in P \rangle$ and the latter is the sought-after symmorphic space group.
1. Step: $f$ is injective
Let be $f([Q_1]) = f([Q_2])$ and write $G_i := G_{Q_i}$. Then there is an
isomorphism $\phi: G_1 \to G_2$ of groups and we wan't to show that
$Q_1, Q_2$ are conjugate in $GL_n(\mathbb{Z})$.
From the lemma in Step 3 it follows that $L_0$ is the unique maximal free abelian rank $n$ subgroup of each $G_Q$. Since maximality is preserved by group isomorphisms,
$\phi(L_0) \cong L_0$ is a maximal free abelian subgroup of rank $n$ of $G_2$
and by uniqueness, $\phi(L_0) = L_0$. Now by a general principal in group
theory, $\phi$ induces an isomorphism $\bar{\phi}: Q_1 \to Q_2$ such that
there is a commutative diagramm:
$$ 1 \to L_0 \to G_1 \overset{\kappa_1}{\to} Q_1 \to 1 $$
$$ \hspace{5pt} \phi \downarrow \hspace{15pt} \phi \downarrow \hspace{17pt} \downarrow\bar{\phi} \hspace{15pt}$$
$$ 1 \to L_0 \to G_2 \underset{\kappa_2}{\to} Q_2 \to 1 $$
$\bar{\phi}$ can be described as follows: to $B \in Q_1$ choose
$X \in G_1$ such that $\kappa_1(X) = B$. Then
$\bar{\phi}(B) = (\kappa_2 \circ \phi)(X)$. (In the current situation we can always take $X=(0,B)$ ).
Since $\phi \in Aut(L_0) = Aut(\mathbb{Z}^n)$, there is
$A \in GL_n(\mathbb{Z})$ such that $\phi(a,I) = (Aa,I)$.
For $B \in Q_1$ write $\phi(0,B) = (b,C)$ with an $C \in Q_2$. Let $e_i$ be a standard vector. From
$$(CAe_i,I)= (b,C) \cdot (Ae_i,I) \cdot (b,C)^{-1} = \phi((0,B) \cdot (e_i,I) \cdot (0,B)^{-1})$$
$$ = \phi(Be_i,I)= (ABe_i,I)\hspace{80pt}$$
one finds $CA=AB$, i.e. $C=ABA^{-1}$. This yields
$$\bar{\phi}(B) = (\kappa_2\circ \phi)(0,B)= \kappa_2(b,ABA^{-1}) = ABA^{-1},$$ hence $\bar{\phi}$
is conjugation with $A$ and $Q_2= \bar{\phi}(Q_1) = AQ_1A^{-1}$ is
conjugated to $Q_1$ in $GL_n(\mathbb{Z})$.
Step 2: $f$ is surjective
Let $G$ be a symmorphic space group. We have to show that there exists
a finite subgroup $Q \le GL_n(\mathbb{Z})$ such that $G \cong G_Q$.
Let $T$ be the translational subgroup of $G$ with lattice
$L = \oplus_{i=1}^n \mathbb{Z}a_i$. Set $A := (a_1,...,a_n) \in GL_n(\mathbb{R})$.
Then $\alpha: L_0 \to L, x \to Ax$ is an isomorphism that induces
a further isomorphism
$$\phi: GL_n(\mathbb{Z}) = Aut(L_0) \to Aut(L), B \mapsto ABA^{-1}.$$
Now let be $P := \mathcal{Q}(L) \le Aut(L) \cap O_n$ (notation in OP in 1.).
In particular, as a discrete subgroup of a compact group, $P$ is finite. Thus $Q := \phi^{-1}(P)$ is a finite subgroup of $GL_n(\mathbb{Z})$ and $\beta: Q \to P, B \mapsto \phi(B) = ABA^{-1}$ is an isomorphism.
Since $G$ is symmorphic, $G = L \rtimes P$ where $P$ acts on $L$ via matrix
multiplication. By definition, $G_Q = L_0 \rtimes Q$ where $Q$ also
operates via matrix multiplication. Now $G \cong G_Q$ follows exactly
as in Step 0.
Step 3: This result was used in Step 1.
Lemma: The translational subgroup is the unique maximal free abelian rank $n$ subgroup of a space group of dimension $n$.
Let $T$ be the translational subgroup of the space group $G$ (dimension $n$) and let $F \le G$ be a free abelian subgroup of rank $n$. From finiteness of $G/T$ follows that $TF/T \cong F/F \cap T$ is finite. Thus $F \cap T$ is free abelian of rank $n$, hence $F \cap T$ has finite index in $T$. Let $t_{a_i} = (a_i,I)$ $(i=1,...,n)$ be basis vectors for $T$. Then we can choose $k \in \mathbb{Z}, k \neq 0$ such that $(t_{a_i})^k = (ka_i,I) \in F$.
Let $(a,A) \in F$. Since $F$ is abelian, $(a,A)$ and $(ka_i,I)$ commute, yielding $(a+Aka_i,A)=(ka_i+a,A)$ and in particular $Aa_i = a_i$. Thus $A=I$, i.e. $(a,A)$ is a translation, which means $(a,A) \in T$. Hence $F \le T$.
Next I show that $T$ is maximal free abelian of rank $n$: Let $F$ be as above and assume $T \le F$. Since $F \le T$ as shown before, $T=F$ follows. Thus $T$ is maximal.
Finally I show that $T$ is the only maximal free abelian subgroup of rank $n$. Let $F$ be another one. Since again $F \le T$, maximality of $F$ implies $F=T.\quad\quad$ q.e.d.
Best Answer
Here's an idea for a proof that the modular group $\Gamma=\mathbb{Z}/2*\mathbb{Z}/3$, which is, of course, virtually free, doesn't surject a group of the form $F\rtimes H$. I don't have time to work out the details.
First, I think it's plausible that the only reduced, non-trivial graph-of-groups decomposition for $\Gamma$ is the one that realises it as the free product given above, which I will denote by $\mathcal{H}$. One idea for the proof is as follows. If $\mathcal{G}$ is any such graph of groups with fundamental group $\Gamma$, then the generators $a$ of order two and $b$ of order 3 can be taken to lie in distinct vertices of $\mathcal{G}$; this leads to an obvious immersion of graphs of groups $\mathcal{H}\to\mathcal{G}$. But $\mathcal{H}$ carries the whole of $\Gamma$, so the immersion is actually an isomorphism.
Remark: If the above is true, then it should be known, and there should be a reference out there somewhere.
Now suppose that $f:\Gamma\to F\rtimes H$ is a surjection. Then $F\rtimes H$ can be decomposed as a graph of groups $\mathcal{J}$, and this induces a graph of groups decomposition $\mathcal{J}'$ for $\Gamma$, by setting $\mathcal{J}'_v=f^{-1}\mathcal{J}_v$ for each vertex $v$ and $\mathcal{J}'_e=f^{-1}\mathcal{J}_e$ for each edge $e$. But, by the previous claim, $\mathcal{J}'=\mathcal{H}$. As $\mathbb{Z}/2$ and $\mathbb{Z}/3$ are both simple, it follows that $f$ is an isomorphism. But $\Gamma$ is not of the form $F\rtimes H$. (To see this, note, for instance, that all the torsion in $F\rtimes H$ is conjugate into a singe finite subgroup $H$; this is not the case in $\Gamma$.)