It seems easier to work this out than to dig up a literature reference. This is an application of the basic structure of connected reductive groups and the dictionary between connected reductive groups and root data. The sketch below  carries out some parts over an arbitrary field whenever possible,
since $F$ could just as well be a global function field or a local field of any characteristic (or whatever).  

For any connected reductive $\mathbf{C}$-group $H$, the group $Z_H(\mathbf{C})$ of $\mathbf{C}$-points of its center $Z_H$ is naturally identified with the group ${\rm{Hom}}({\rm{X}}(Z_H), \mathbf{C}^{\times})$ of homomorphisms from its own character group ${\rm{X}}(Z_H) := {\rm{Hom}}(Z_H, \mathbf{G}_m)$ into $\mathbf{C}^{\times}$. Hence, it is equivalent to prove short-exactness of the induced diagram between the character groups of the centers of the dual groups.  

For any split connected reductive group $H$ over a field $k$, with associated root datum $(X, \Phi, X^{\vee}, \Phi^{\vee})$ (relative to a choice of split maximal $k$-torus $T \subset H$ and Borel $k$-subgroup $B \subset H$ containing $T$), the geometric character group of the center of $H$ is $X/\mathbf{Z}\Phi$ (e.g., $Z_H=1$, which is to say $H$ is of adjoint type, precisely when $\Phi$ spans $X$ over $\mathbf{Z}$).  

Recalling how the root datum of the dual group is designed via the dual root datum, the task then is to prove short-exactness of the diagram
$$0 \rightarrow X_1^{\vee}/\mathbf{Z}\Phi_1^{\vee} \rightarrow X_2^{\vee}/\mathbf{Z}\Phi_2^{\vee} \rightarrow X_3^{\vee}/\mathbf{Z}\Phi_3^{\vee} \rightarrow 0$$
(where we make the root data relative to a choice of split maximal torus $T_2 \subset (G_2)_{F_s}$, Borel $F_s$-subgroup $B_2 \subset (G_2)_{F_s}$ containing $T_2$, the associated maximal tori $T_1 = (G_1)_{F_s} \cap T_2 \subset (G_1)_{F_s}$ and $T_3 = {\rm{im}}(T_2 \rightarrow (G_3)_{F_s}) \subset (G_3)_{F_s}$, and the associated Borel $F_s$-subgroups $B_1 = (G_1)_{F_s} \cap B_2 \subset (G_1)_{F_s}$
and $B_3 = {\rm{im}}(B_2 \rightarrow (G_3)_{F_s}) \subset (G_3)_{F_s}$ containing $T_1$ and $T_3$ respectively).  Of course, one has to prove that such $T_1, B_1, T_3, B_3$ really are as advertised (no disconnectedness ambiguity!), especially $T_1$ and $B_2$; this is left as an **exercise** using the normality of $G_1$ in $G_2$.

Since $T_3 = T_2/T_1$ with every $T_i$ a torus, it is clear that the induced complex of cocharacter lattices
$$0 \rightarrow X_1^{\vee} \rightarrow X_2^{\vee} \rightarrow X_3^{\vee} \rightarrow 0$$
is short exact. Hence, it remains to prove two things:

(i) the evident complex
$$0 \rightarrow \mathbf{Z}\Phi_1^{\vee} \rightarrow \mathbf{Z}\Phi_2^{\vee} \rightarrow \mathbf{Z}\Phi_3^{\vee} \rightarrow 0$$
is short exact, 


(ii)  the evident inclusion
$$\mathbf{Z}\Phi_1^{\vee} \subset X_1^{\vee} \cap \mathbf{Z}\Phi_2^{\vee}$$
is an equality. 

We'll see that (i) is a consequence of properties of root systems, and (ii) ultimately reduces to (i) by using
facts about simply connected semisimple groups.

Recall that any root system is uniquely a direct sum of irreducible components.  Moreover, 
the irreducible components of a split connected reductive group are naturally identified with with the root systems of the simple
connected semisimple normal subgroups (**exercise**).  Hence, $\Phi_2$ is naturally identified with
the *direct sum* of the root systems $\Phi_1$ and $\Phi_3$.  From this one obtains (i) immediately. 

The proof of (ii) lies a bit deeper, and involves
the simply connected central covers of the derived groups.  More precisely, let the central isogeny $$q_i :\widetilde{G}_i \rightarrow \mathscr{D}(G_i)$$ be the simply connected central cover of the derived group of $G_i$, and let $\widetilde{T}_i := (q_i)_{F_s}^{-1}(T'_i)$ be the maximal $F_s$-torus preimage of the maximal $F_s$-torus  $T'_i := T_i \cap \mathscr{D}(G_i)_{F_s}$ in $\mathscr{D}(G_i)_{F_s} = \mathscr{D}((G_i)_{F_s})$.
By the standard dictionary between root data and connected reductive groups,
$\mathbf{Z}\Phi_i^{\vee}$ is the image of the injective map ${\rm{X}}_{\ast}(\widetilde{T}_i) \hookrightarrow {\rm{X}}_{\ast}(T_i) =
X_i^{\vee}$ between cocharacter lattices (injective since $\widetilde{T}_i \rightarrow T_i$ has finite kernel).  

To exploit this link to maximal tori in the simply connected central cover of the derived group, we shall use
the functoriality of the formation of the simply connected central cover of a connected semisimple group.
More precisely:

**Theorem**: *For any field $k$ and $k$-homomorphism
$f:H \rightarrow H'$  between connected semisimple $k$-groups, 
there is a unique compatible $k$-homomorphism $\widetilde{f}:\widetilde{H} \rightarrow \widetilde{H}'$ between the respective 
simply connected central covers $q: \widetilde{H} \rightarrow H$ and $q': \widetilde{H}' \rightarrow H'$.*

We'll prove this Theorem (inspired by topology) at the end; we only truly need it for separably closed $k$ (but we'll apply it over $F$ because that makes us feel good). 

Returning to our situation of interest, we get a unique complex of connected semisimple $F$-groups
$$1 \rightarrow \widetilde{G}_1 \rightarrow \widetilde{G}_2 \rightarrow \widetilde{G}_3 \rightarrow 1$$
over the induced diagram of derived groups, and this induces a diagram of maximal $F_s$-tori
$$\widetilde{T}_1 \rightarrow \widetilde{T}_2 \rightarrow \widetilde{T}_3.$$
Our aim is to prove that the inclusion 
${\rm{X}}_{\ast}(\widetilde{T}_1) \subset {\rm{X}}_{\ast}(T_1) \cap {\rm{X}}_{\ast}(\widetilde{T}_2)$
inside ${\rm{X}}_{\ast}(T_2)$ is an equality.
Since $\widetilde{T}_1 := (q_1)_{F_s}^{-1}(T_1)$,  we just need to check that 
any cocharacter $\lambda:\mathbf{G}_m \rightarrow \widetilde{T}_2$ for which 
$(q_2)_{F_s} \circ \lambda: \mathbf{G}_m \rightarrow T_2$ lands inside $T_1$ must itself 
factor through the $F_s$-homomorphism $\widetilde{T}_1 \rightarrow T_1$
whose kernel is finite.   (An equivalent assertion is that the map of tori $\widetilde{T}_1 \rightarrow
(T_1 \times_{T_2} \widetilde{T}_2)^0_{\rm{red}}$ is an isomorphism.)

Since $T_1 = T_2 \cap (G_1)_{F_s}$, $T_3 = T_2/T_1$, $G_3 = G_2/G_1$, and
$\widetilde{G}_3 \rightarrow G_3$ has finite kernel, 
it is equivalent to show that if $\lambda$ is killed by $(\widetilde{G}_2)_{F_s} \rightarrow
(\widetilde{G}_3)_{F_s}$ then $\lambda$ factors (necessarily uniquely) through $(\widetilde{G}_1)_{F_s} \rightarrow
(\widetilde{G}_2)_{F_s}$.  Better yet, we claim that the complex of connected semisimple groups
$$1 \rightarrow \widetilde{G}_1 \rightarrow \widetilde{G}_2 \rightarrow \widetilde{G}_3 \rightarrow 1$$
is a short exact sequence that is even uniquely split as a direct product.  This can be shown in various ways,
but one has to be careful not to engage in circular reasoning (i.e., one has to remember whatever foundational
route one took when learning the structure theory of connected reductive groups).
We offer an argument via reduction to the assertion (i) above. 

Letting 
$\widetilde{X}_i = {\rm{X}}(\widetilde{T}_i)$, 
by the Isomorphism Theorem it is equivalent to show that the semisimple root datum
$(\widetilde{X}_2, \Phi_2)$ is the direct product of the semisimple root data 
$(\widetilde{X}_1, \Phi_1)$ and $(\widetilde{X}_3, \Phi_3)$ compatibly with the known 
decomposition of $\Phi_2$ as the direct sum of the root
systems $\Phi_1$ and $\Phi_3$.   This in turn amounts to the short exactness of the complex 
$$0 \rightarrow \widetilde{X}_3 \rightarrow \widetilde{X}_2 \rightarrow \widetilde{X}_1 \rightarrow 0.$$
But $\widetilde{X}_i$ is the $\mathbf{Z}$-dual of
$\mathbf{Z}\Phi_i^{\vee}$, so by (i) we are done!

Finally, it remans to give:

Proof of Theorem:  The pullback $k$-group scheme
$$E := \widetilde{H} \times_{H'} \widetilde{H}'$$
is a central extension of $\widetilde{H}$ by the finite group scheme $\mu' := \ker(\widetilde{H}' \rightarrow H')$, and our task is to prove that this central extension
$$1 \rightarrow \mu' \rightarrow E \rightarrow \widetilde{H} \rightarrow 1$$
is uniquely split.  (It is a general fact that any central extension of a simply connected semisimple
group by a group scheme with no nontrivial smooth connected subgroup is uniquely split,
but we will give a direct argument in our situation.) 

Note that $E$ might not be $k$-smooth if ${\rm{char}}(k) > 0$,
and perhaps $k$ is not perfect.  But no worries:
let $S' \subset \widetilde{H}'$ be a maximal $k$-torus, so $\mu' \subset S'$ (as $\mu' \subset Z_{\widetilde{H}'} 
\subset Z_{\widetilde{H}'}(S') = S'$), and form the pushout central extension
$$1 \rightarrow S' \rightarrow \mathscr{E} = S' \times^{\mu'} E \rightarrow \widetilde{H} \rightarrow 1.$$
Since $\mathscr{E}$ is a central extension of $\widetilde{H}$ by a $k$-torus, clearly $\mathscr{E}$
is smooth, connected, and even reductive.  Hence, $\mathscr{D}(\mathscr{E})$ makes sense (!)
and is a connected semisimple $k$-group that has finite center and hence finite intersection with the central torus $S'$.
In other words, the evident surjection $\mathscr{D}(\mathscr{E}) \rightarrow \mathscr{D}(\widetilde{H}) =
\widetilde{H}$ is a *central isogeny* between connected semisimple groups.  

Aha, but
$\widetilde{H}$ is *simply connected*, so consideration of root data over $k_s$ 
(thinking back to how simple connectedness is actually defined for connected semisimple groups) shows that $\widetilde{H}$
has no nontrivial connected semisimple central extension. Hence, $\mathscr{D}(\mathscr{E}) \rightarrow
\widetilde{H}$ is an isomorphism.  Since the connected reductive $\mathscr{E}$ is generated by
its commuting $k$-subgroups $S'$ and $\mathscr{D}(\widetilde{E})$, the inverse isomorphism
$\widetilde{H} \simeq  \mathscr{D}(\mathscr{E})$ thereby defines a splitting 
of $\mathscr{E}$ as a central extension; this is even unique since
${\rm{Hom}}_k(\widetilde{H}, S') = 1$ due to perfectness of $\widetilde{H}$ and commutativity
of $S'$.  

From the construction of $\mathscr{E}$
as a central pushout we see that $E \subset \mathscr{E}$ with commutative cokernel, so
the resulting inclusion $\widetilde{H} = \mathscr{D}(\mathscr{E}) \subset \mathscr{E}$ must
factor through $E$ (as the perfect smooth affine $k$-group $\widetilde{H}$ has no nontrivial
$k$-homomorphism to a commutative $k$-group scheme, such as $\mathscr{E}/E$).  This provides a splitting
of $E$ as a central extension, and it is unique since ${\rm{Hom}}_k(\widetilde{H}, \mu') = 1$.
QED