Firstly, there's no guarantee that composing two symplectic transformations with $q-q$ generating function will produce another symplectic transformation with a $q-q$ generating function (see below for an example), so some conditions need to be true for this to happen, which will emerge in the discussion below.

By definition of generating functions, $F_1$ satisfies $$ p_1\cdot dq^1-p_2\cdot dq^2 = \nabla_{q^1}F_1\cdot dq^1 + \nabla_{q^2}F_1\cdot dq^2 $$ while $F_2$ satisfies $$ p_2\cdot dq^2-p_3\cdot dq^3 = \nabla_{q^2}F_2\cdot dq^2 + \nabla_{q^3}F_2\cdot dq^3 $$ Adding these equations yields $$ p_1\cdot dq^1 - p_3\cdot dq^3 = \nabla_{q^1}F_1\cdot dq^1 + \nabla_{q^2}(F_1+F_2)\cdot dq^2 + \nabla_{q^3}F_2\cdot dq^3 $$ Both sides here are considered as functions of general $q_1,q_2,q_3$. To produce a generating function of $T=T_2T_1$, we need to pick a function $q^2 = {q^2}^*(q^1,q^3)$ so that the middle term on the RHS vanishes. In other words, the equation $\nabla_{q^2}(F_1+F_2) = 0$ needs to have a well-defined solution ${q^2}^*(q^1,q^3)$. By the implicit function theorem, this will be possible (at least locally) if the Hessian of $q^2 \mapsto F_1(q^1,q^2)+F_2(q^2,q^3)$ is invertible for all $q^1,q^3$.

Assuming this is the case, the generating function for $T$ will be $$ F(q^1,q^3) := F_1(q^1,{q^2}^*(q^1,q^3)) + F_2({q^2}^*(q^1,q^3),q^3). $$

As an example of a case where this doesn't work, consider the generating functions $F_1(q^1,q^2) = q^1q^2$ and $F_2(q^2,q^3) = q^2q^3$ (where for simplicity I'm just considering two-dimensional phase spaces). The corresponding symplectomorphisms are $$ T_1(q^1,p_1) = (p_1, -q^1), \quad T_2(q^2,p_2) = (p_2, -q^2), \quad T(q^1,p_1) = (-q^1,-p_1). $$ The composition $T$ is not generated by a function of the form $F(q^1,q^3)$. The above proof fails since the Hessian of $q^2 \mapsto q^1q^2+q^2q^3$ is zero.

Edit: I'm realising that while the above gives necessary conditions for constructing the generating function of the composition, these conditions may not be sufficient. For example, consider $$ F_1(q^1,q^2) = q^1(q^2)^2,\quad F_2(q^2,q^3) = (q^2)^2q^3 $$ which generate the symplectomorphisms $$ T_1(q^1,p_1) = (\pm\sqrt{p_1}, \mp 2q^1\sqrt{p_1}),\quad T_2(q^2,p_2) = (p_2/(2q^2), - (q^2)^2). $$ The composition is again $T(q^1,p_1) = (-q^1,-p^1)$. The construction above requires solving $$ \nabla_{q^2}(F_1+F_2) = 2q^2(q^1+q^3) = 0 \ \forall q^1,q^3 \implies {q^2}^*(q^1,q^3) = 0 $$ and hence $$ F(q^1,q^3) = F_1(q^1,0) + F_2(0,q^3) = 0, $$ which isn't a proper generating function. So it seems there also needs to be some requirement that the constructed $F$ actually defines a transformation.

Firstly, there's no guarantee that composing two symplectic transformations with $q-q$ generating function will produce another symplectic transformation with a $q-q$ generating function (see below for an example), so some conditions need to be true for this to happen, which will emerge in the discussion below.

By definition of generating functions, $F_1$ satisfies $$ p_1\cdot dq^1-p_2\cdot dq^2 = \nabla_{q^1}F_1\cdot dq^1 + \nabla_{q^2}F_1\cdot dq^2 $$ while $F_2$ satisfies $$ p_2\cdot dq^2-p_3\cdot dq^3 = \nabla_{q^2}F_2\cdot dq^2 + \nabla_{q^3}F_2\cdot dq^3 $$ Adding these equations yields $$ p_1\cdot dq^1 - p_3\cdot dq^3 = \nabla_{q^1}F_1\cdot dq^1 + \nabla_{q^2}(F_1+F_2)\cdot dq^2 + \nabla_{q^3}F_2\cdot dq^3 $$ Both sides here are considered as functions of general $q_1,q_2,q_3$. To produce a generating function of $T=T_2T_1$, we need to pick a function $q^2 = {q^2}^*(q^1,q^3)$ so that the middle term on the RHS vanishes. In other words, the equation $\nabla_{q^2}(F_1+F_2) = 0$ needs to have a well-defined solution ${q^2}^*(q^1,q^3)$. By the implicit function theorem, this will be possible (at least locally) if the Hessian of $q^2 \mapsto F_1(q^1,q^2)+F_2(q^2,q^3)$ is invertible for all $q^1,q^3$.

Assuming this is the case, the generating function for $T$ will be $$ F(q^1,q^3) := F_1(q^1,{q^2}^*(q^1,q^3)) + F_2({q^2}^*(q^1,q^3),q^3). $$

As an example of a case where this doesn't work, consider the generating functions $F_1(q^1,q^2) = q^1q^2$ and $F_2(q^2,q^3) = q^2q^3$ (where for simplicity I'm just considering two-dimensional phase spaces). The corresponding symplectomorphisms are $$ T_1(q^1,p_1) = (p_1, -q^1), \quad T_2(q^2,p_2) = (p_2, -q^2), \quad T(q^1,p_1) = (-q^1,-p_1). $$ The composition $T$ is not generated by a function of the form $F(q^1,q^3)$. The above proof fails since the Hessian of $q^2 \mapsto q^1q^2+q^2q^3$ is zero.

Edit: I'm realising that while the above gives necessary conditions for constructing the generating function of the composition, these conditions may not be sufficient. For example, consider $$ F_1(q^1,q^2) = q^1(q^2)^2,\quad F_2(q^2,q^3) = (q^2)^2q^3 $$ which generate the symplectomorphisms $$ T_1(q^1,p_1) = (\pm\sqrt{p_1}, \mp 2q^1\sqrt{p_1}),\quad T_2(q^2,p_2) = (p_2/(2q^2), - (q^2)^2). $$ The composition is again $T(q^1,p_1) = (-q^1,-p_1)$. The construction above requires solving $$ \nabla_{q^2}(F_1+F_2) = 2q^2(q^1+q^3) = 0 \ \forall q^1,q^3 \implies {q^2}^*(q^1,q^3) = 0 $$ and hence $$ F(q^1,q^3) = F_1(q^1,0) + F_2(0,q^3) = 0, $$ which isn't a proper generating function. So it seems there also needs to be some requirement that the constructed $F$ actually defines a transformation.

Firstly, there's no guarantee that composing two symplectic transformations with $q-q$ generating function will produce another symplectic transformation with a $q-q$ generating function (see below for an example), so some conditions need to be true for this to happen, which will emerge in the discussion below.

By definition of generating functions, $F_1$ satisfies $$ p_1\cdot dq^1-p_2\cdot dq^2 = \nabla_{q^1}F_1\cdot dq^1 + \nabla_{q^2}F_1\cdot dq^2 $$ while $F_2$ satisfies $$ p_2\cdot dq^2-p_3\cdot dq^3 = \nabla_{q^2}F_2\cdot dq^2 + \nabla_{q^3}F_2\cdot dq^3 $$ Adding these equations yields $$ p_1\cdot dq^1 - p_3\cdot dq^3 = \nabla_{q^1}F_1\cdot dq^1 + \nabla_{q^2}(F_1+F_2)\cdot dq^2 + \nabla_{q^3}F_2\cdot dq^3 $$ Both sides here are considered as functions of general $q_1,q_2,q_3$. To produce a generating function of $T=T_2T_1$, we need to pick a function $q^2 = {q^2}^*(q^1,q^3)$ so that the middle term on the RHS vanishes. In other words, the equation $\nabla_{q^2}(F_1+F_2) = 0$ needs to have a well-defined solution ${q^2}^*(q^1,q^3)$. By the implicit function theorem, this will be possible (at least locally) if the Hessian of $q^2 \mapsto F_1(q^1,q^2)+F_2(q^2,q^3)$ is invertible for all $q^1,q^3$.

Assuming this is the case, the generating function for $T$ will be $$ F(q^1,q^3) := F_1(q^1,{q^2}^*(q^1,q^3)) + F_2({q^2}^*(q^1,q^3),q^3). $$

As an example of a case where this doesn't work, consider the generating functions $F_1(q^1,q^2) = q^1q^2$ and $F_2(q^2,q^3) = q^2q^3$ (where for simplicity I'm just considering two-dimensional phase spaces). The corresponding symplectomorphisms are $$ T_1(q^1,p_1) = (p_1, -q^1), \quad T_2(q^2,p_2) = (p_2, -q^2), \quad T(q^1,p_1) = (-q^1,-p_1). $$ The composition $T$ is not generated by a function of the form $F(q^1,q^3)$. The above proof fails since the Hessian of $q^2 \mapsto q^1q^2+q^2q^3$ is zero.

Firstly, there's no guarantee that composing two symplectic transformations with $q-q$ generating function will produce another symplectic transformation with a $q-q$ generating function (see below for an example), so some conditions need to be true for this to happen, which will emerge in the discussion below.

By definition of generating functions, $F_1$ satisfies $$ p_1\cdot dq^1-p_2\cdot dq^2 = \nabla_{q^1}F_1\cdot dq^1 + \nabla_{q^2}F_1\cdot dq^2 $$ while $F_2$ satisfies $$ p_2\cdot dq^2-p_3\cdot dq^3 = \nabla_{q^2}F_2\cdot dq^2 + \nabla_{q^3}F_2\cdot dq^3 $$ Adding these equations yields $$ p_1\cdot dq^1 - p_3\cdot dq^3 = \nabla_{q^1}F_1\cdot dq^1 + \nabla_{q^2}(F_1+F_2)\cdot dq^2 + \nabla_{q^3}F_2\cdot dq^3 $$ Both sides here are considered as functions of general $q_1,q_2,q_3$. To produce a generating function of $T=T_2T_1$, we need to pick a function $q^2 = {q^2}^*(q^1,q^3)$ so that the middle term on the RHS vanishes. In other words, the equation $\nabla_{q^2}(F_1+F_2) = 0$ needs to have a well-defined solution ${q^2}^*(q^1,q^3)$. By the implicit function theorem, this will be possible (at least locally) if the Hessian of $q^2 \mapsto F_1(q^1,q^2)+F_2(q^2,q^3)$ is invertible for all $q^1,q^3$.

Assuming this is the case, the generating function for $T$ will be $$ F(q^1,q^3) := F_1(q^1,{q^2}^*(q^1,q^3)) + F_2({q^2}^*(q^1,q^3),q^3). $$

As an example of a case where this doesn't work, consider the generating functions $F_1(q^1,q^2) = q^1q^2$ and $F_2(q^2,q^3) = q^2q^3$ (where for simplicity I'm just considering two-dimensional phase spaces). The corresponding symplectomorphisms are $$ T_1(q^1,p_1) = (p_1, -q^1), \quad T_2(q^2,p_2) = (p_2, -q^2), \quad T(q^1,p_1) = (-q^1,-p_1). $$ The composition $T$ is not generated by a function of the form $F(q^1,q^3)$. The above proof fails since the Hessian of $q^2 \mapsto q^1q^2+q^2q^3$ is zero.

Edit: I'm realising that while the above gives necessary conditions for constructing the generating function of the composition, these conditions may not be sufficient. For example, consider $$ F_1(q^1,q^2) = q^1(q^2)^2,\quad F_2(q^2,q^3) = (q^2)^2q^3 $$ which generate the symplectomorphisms $$ T_1(q^1,p_1) = (\pm\sqrt{p_1}, \mp 2q^1\sqrt{p_1}),\quad T_2(q^2,p_2) = (p_2/(2q^2), - (q^2)^2). $$ The composition is again $T(q^1,p_1) = (-q^1,-p^1)$. The construction above requires solving $$ \nabla_{q^2}(F_1+F_2) = 2q^2(q^1+q^3) = 0 \ \forall q^1,q^3 \implies {q^2}^*(q^1,q^3) = 0 $$ and hence $$ F(q^1,q^3) = F_1(q^1,0) + F_2(0,q^3) = 0, $$ which isn't a proper generating function. So it seems there also needs to be some requirement that the constructed $F$ actually defines a transformation.