I am not sure I really understand the question, but: suppose $k=2,$ and our maps are $\phi_1, \phi_2.$ Then the composition of your maps is $\phi: x -> \alpha_1 \alpha_2 x + \alpha_2 \beta_1 + \beta_2.$ Now, it is not too hard to recover the two numbers $\rho = \alpha_1 \alpha_2$ and $\sigma = \alpha_2 \beta_1 + \beta_2$ from the image of the integers under $\phi.$ But now, the set $\alpha_1 \alpha_2 = \rho; \alpha_2 \beta_1 + \beta_2 = \sigma$ is an (affine) two dimensional variety in $\mathbb{R}^4,$ so there will be a point with coordinates $\alpha_1^\prime, \alpha_2^\prime, \beta_1^\prime, \beta_2^\prime$ which are not algebraically related to the original $\alpha_1, \alpha_2, \beta_1, \beta_2,$ so picking $\phi_3, \phi_4$ with the new $\alpha, \beta$ we obtain a counterexample for $k=4.$ Of course, I am not sure what the OP means by "mutually algebraically independent", I would assume that means that for no pair $x_1, x_2$ (where $x$es are one of $\alpha_i, \beta_j$) is there a bivariate polynomial $ \in \mathbb{Z}[x, y]$ which vanishes for $x=x_1, y=x_2.$

More thoughts

In fact, the problem has a number of aspects?

Does knowing the action on $\mathbb{Z}$ imply knowing the action?

In general, of course the answer is NO, since while the "rotational" part of the action can be computed from the action on the integers (the set of differences of values at $\mathbb{Z}$ will be a lattice in $\mathbb{R}$), but the translational part can only be determined up to integer multiple of some magic quantity (that is, transformations $\phi_k=\alpha(z + \beta + k)$ will have the same effect on the integers for all integers $k). However, with the given set of components, as in the OP's question, being able to get two different $\phi_k$ would imply an algebraic relation. BUT, that relation might be a consequence of the group laws.

Now, suppose we do know the product map. Do we know that there is a unique way to get it? As @MichaelZieve points out, the answer is no, and since we are in an algebraic group, the group laws are the only obstructions.

Finally, is the commutation of commutators the only law in the affine group? Since this is true in any two-step solvable group, this looks a little suspicious.

I am not sure I really understand the question, but: suppose $k=2,$ and our maps are $\phi_1, \phi_2.$ Then the composition of your maps is $\phi: x -> \alpha_1 \alpha_2 x + \alpha_2 \beta_1 + \beta_2.$ Now, it is not too hard to recover the two numbers $\rho = \alpha_1 \alpha_2$ and $\sigma = \alpha_2 \beta_1 + \beta_2$ from the image of the integers under $\phi.$ But now, the set $\alpha_1 \alpha_2 = \rho; \alpha_2 \beta_1 + \beta_2 = \sigma$ is an (affine) two dimensional variety in $\mathbb{R}^4,$ so there will be a point with coordinates $\alpha_1^\prime, \alpha_2^\prime, \beta_1^\prime, \beta_2^\prime$ which are not algebraically related to the original $\alpha_1, \alpha_2, \beta_1, \beta_2,$ so picking $\phi_3, \phi_4$ with the new $\alpha, \beta$ we obtain a counterexample for $k=4.$ Of course, I am not sure what the OP means by "mutually algebraically independent", I would assume that means that for no pair $x_1, x_2$ (where $x$es are one of $\alpha_i, \beta_j$) is there a bivariate polynomial $ \in \mathbb{Z}[x, y]$ which vanishes for $x=x_1, y=x_2.$

More thoughts

In fact, the problem has a number of aspects?

Does knowing the action on $\mathbb{Z}$ imply knowing the action?

In general, of course the answer is NO, since while the "rotational" part of the action can be computed from the action on the integers (the set of differences of values at $\mathbb{Z}$ will be a lattice in $\mathbb{R}$), but the translational part can only be determined up to integer multiple of some magic quantity (that is, transformations $\phi_k=\alpha(z + \beta + k)$ will have the same effect on the integers for all integers $k$). However, with the given set of components, as in the OP's question, being able to get two different $\phi_k$ would imply an algebraic relation. BUT, that relation might be a consequence of the group laws.

Now, suppose we do know the product map. Do we know that there is a unique way to get it? As @MichaelZieve points out, the answer is no, and since we are in an algebraic group, the group laws are the only obstructions.

Finally, is the commutation of commutators the only law in the affine group? Since this is true in any two-step solvable group, this looks a little suspicious.

I am not sure I really understand the question, but: suppose $k=2,$ and our maps are $\phi_1, \phi_2.$ Then the composition of your maps is $\phi: x -> \alpha_1 \alpha_2 x + \alpha_2 \beta_1 + \beta_2.$ Now, it is not too hard to recover the two numbers $\rho = \alpha_1 \alpha_2$ and $\sigma = \alpha_2 \beta_1 + \beta_2$ from the image of the integers under $\phi.$ But now, the set $\alpha_1 \alpha_2 = \rho; \alpha_2 \beta_1 + \beta_2 = \sigma$ is an (affine) two dimensional variety in $\mathbb{R}^4,$ so there will be a point with coordinates $\alpha_1^\prime, \alpha_2^\prime, \beta_1^\prime, \beta_2^\prime$ which are not algebraically related to the original $\alpha_1, \alpha_2, \beta_1, \beta_2,$ so picking $\phi_3, \phi_4$ with the new $\alpha, \beta$ we obtain a counterexample for $k=4.$ Of course, I am not sure what the OP means by "mutually algebraically independent", I would assume that means that for no pair $x_1, x_2$ (where $x$es are one of $\alpha_i, \beta_j$) is there a bivariate polynomial $ \in \mathbb{Z}[x, y]$ which vanishes for $x=x_1, y=x_2.$

I am not sure I really understand the question, but: suppose $k=2,$ and our maps are $\phi_1, \phi_2.$ Then the composition of your maps is $\phi: x -> \alpha_1 \alpha_2 x + \alpha_2 \beta_1 + \beta_2.$ Now, it is not too hard to recover the two numbers $\rho = \alpha_1 \alpha_2$ and $\sigma = \alpha_2 \beta_1 + \beta_2$ from the image of the integers under $\phi.$ But now, the set $\alpha_1 \alpha_2 = \rho; \alpha_2 \beta_1 + \beta_2 = \sigma$ is an (affine) two dimensional variety in $\mathbb{R}^4,$ so there will be a point with coordinates $\alpha_1^\prime, \alpha_2^\prime, \beta_1^\prime, \beta_2^\prime$ which are not algebraically related to the original $\alpha_1, \alpha_2, \beta_1, \beta_2,$ so picking $\phi_3, \phi_4$ with the new $\alpha, \beta$ we obtain a counterexample for $k=4.$ Of course, I am not sure what the OP means by "mutually algebraically independent", I would assume that means that for no pair $x_1, x_2$ (where $x$es are one of $\alpha_i, \beta_j$) is there a bivariate polynomial $ \in \mathbb{Z}[x, y]$ which vanishes for $x=x_1, y=x_2.$

More thoughts

In fact, the problem has a number of aspects?

Does knowing the action on $\mathbb{Z}$ imply knowing the action?

In general, of course the answer is NO, since while the "rotational" part of the action can be computed from the action on the integers (the set of differences of values at $\mathbb{Z}$ will be a lattice in $\mathbb{R}$), but the translational part can only be determined up to integer multiple of some magic quantity (that is, transformations $\phi_k=\alpha(z + \beta + k)$ will have the same effect on the integers for all integers $k). However, with the given set of components, as in the OP's question, being able to get two different $\phi_k$ would imply an algebraic relation. BUT, that relation might be a consequence of the group laws.

Now, suppose we do know the product map. Do we know that there is a unique way to get it? As @MichaelZieve points out, the answer is no, and since we are in an algebraic group, the group laws are the only obstructions.

Finally, is the commutation of commutators the only law in the affine group? Since this is true in any two-step solvable group, this looks a little suspicious.