5 added 1 characters in body

This is not an answer to the questions but some general comments. One should be aware that the relation between vector fields and Hasse derivations in characteristic $p$ is not at all analogous to the characteristic $0$. It is true that a Hasse derivation in all characteristics is the same thing as an acction of the formal additive group. The difference is whereas in characteristic $0$ the formal additive group is the only $1$-dimensional formal group whereas in positive characteristic there are more, for instance the formal multiplicative group. In positive characteristic the derivation part of a Hasse derivation corresponds to an action of the group scheme that is the kernel of the Frobenius map which is $\alpha_p$. Again there are several group schemes of order $p$ (such as $\mu_p$ which is the kernel of Frobenius on the formal multiplicative group). On the vector field side there thus is a first obstruction on the vector field itself, that it should give rise to an $\alpha_p$-action. Concretely that means that the $p$-th power of the derivation should be $0$.

Note that that means that there may not even be a vector field to start with even though there might be many vector fields on $M$ there may not be any with that property (for instance a smooth proper toric variety with no automorphisms outside of the torus).

In any case if one wants an obstruction theory one should note that given $D_1$, $D_n$ for $n=1,\ldots,p-1$ is just $D_1^n$ so the first undetermined one would be $D_p$. If one has local liftings one should compare two such liftings $D_p$ and $D'_p$ and it follows that their difference $D'_p-D_p$ is a derivation and one gets a torsor of the tangent sheaf as a first obstruction. Unfortonately if $D$ is a derivation $D'_p=D_p+D$ may not fulfil the further condition for being a part of a Hasse derivation namely that its $p$'th power should be zero. One can expand its $p$'th power using the Jacobson formula but it leads to a (seemingly) nasty non-linear problem. Anyway if solvable one can continuer with $D_{p^2}$ which would be the next undetermined term and continue in the same manner but it looks like a nightmare (the fact that I don't think that local liftings may exist makes it even less palatable).

However, it is clear that you do not have unique extensions: Take some $M$ on which an action of $\hat G_a\times \hat G_a$ is given. Then the actions of $\hat G_a$ given by the inclusions $t\mapsto (t,0)$ and $t\mapsto (t,t^p)$ have the same first order action.

Addendum: I think that all in all Hasse-Schmidt derivations are better than Hasse derivations (recall that a Hasse-Schmidt derivation is just a map $D_\infty\times M\to M$, i.e., not necessarily a formal group action). There the liftings of an order $n$ HS-derivation to order $n+1$ is just a pseudo-torsor over the tangent sheaf with local liftings existing so that the obstruction for extension is just an $H^1$. Of course you will have many more of them but as you don't have uniqueness for lifting vector fields to Hasse derivations anyway it seems to matter less.

Addendum (formal multiplicative group): Whereas actions of the formal additive group are quite messy, the formal multiplicative group (as always with tori) is much nicer. As $\mu_p$ is linearly reductive an action of it is nothing but a $\mathbb Z_p$ action Z/p$-grading on$M$. This is the best description though in terms of vector fields it corresponds to a derivation$D$with$D^p=D$. Similarly a$\mu_{p^n}$-action is nothing but a$\mathbb Z/p^n$-grading. Hence, lifting from$\mu_{p^n}$to$\mu_{p^{n+1}}$corresponds to a refinement of a$\mathbb Z/p^n$-grading to a$\mathbb Z/p^{n+1}$-grading. Hence an action of the formal multiplicative group is the same thing as a$\mathbb Z_p$-grading. NB: As a topological group that is; a collection of compatible$\mathbb Z/p^n$-gradings for each$n$. In particular the part of degree$a\in \mathbb Z_p$is an infinite intersection and could very well be$0$for all$a$. On the other hand an action of the actual multiplicative group is the same thing as a$\mathbb Z$-grading and the associated$\mathbb Z_p$-grading is just induced by$\mathbb Z\subset \mathbb Z_p$. However things can be more complicated yet very close to such an action. If we have a derivation$D$with$D^p=fD$where$f$is an invertible function then we don't have a$\mu_p$-action unless$f=1$and we may not even be able to get a$\mu_p$-action by scaling$D$. However, on the étale cover where we extract a$p-1$'th root of$f$we do get such a modification. So what we have is a kind of twisted$\mu_p$-action and a sheaf of gradings which is an abelian sheaf of$M$isomorphic to$\mathbb Z/p$over the above étale cover (in fact it is the twist of$\mathbb Z/p$by exactly this cover). This extends without problem to twisted actions of the formal multiplicative group. Such thingies do occur in practice. My favourite example is the following: Look at the moduli space (i.e., stack if you want to quibble) of ordinary elliptic curves. Formally at each point this space can be identified using Serre-Tate coordinates (which I guess in this$1$-dimensional case is earlier than Serre-Tate) with the formal multiplicative group. We can now consider the (pro)-étale covering (with structure group$\mathbb Z_p^\ast$) trivialising the group of$p$-torsion points of the universal elliptic curve. By Igusa this is connected and on it we have an action of the formal multiplicative group realising formally at each point the Serre-Tate coordinates. 4 Added details on formal multiplicative group actions Addendum (formal multiplicative group): Whereas actions of the formal additive group are quite messy, the formal multiplicative group (as always with tori) is much nicer. As$\mu_p$is linearly reductive an action of it is nothing but a$\mathbb Z_p$action on$M$. This is the best description though in terms of vector fields it corresponds to a derivation$D$with$D^p=D$. Similarly a$\mu_{p^n}$-action is nothing but a$\mathbb Z/p^n$-grading. Hence, lifting from$\mu_{p^n}$to$\mu_{p^{n+1}}$corresponds to a refinement of a$\mathbb Z/p^n$-grading to a$\mathbb Z/p^{n+1}$-grading. Hence an action of the formal multiplicative group is the same thing as a$\mathbb Z_p$-grading. NB: As a topological group that is; a collection of compatible$\mathbb Z/p^n$-gradings for each$n$. In particular the part of degree$a\in \mathbb Z_p$is an infinite intersection and could very well be$0$for all$a$. On the other hand an action of the actual multiplicative group is the same thing as a$\mathbb Z$-grading and the associated$\mathbb Z_p$-grading is just induced by$\mathbb Z\subset \mathbb Z_p$. However things can be more complicated yet very close to such an action. If we have a derivation$D$with$D^p=fD$where$f$is an invertible function then we don't have a$\mu_p$-action unless$f=1$and we may not even be able to get a$\mu_p$-action by scaling$D$. However, on the étale cover where we extract a$p-1$'th root of$f$we do get such a modification. So what we have is a kind of twisted$\mu_p$-action and a sheaf of gradings which is an abelian sheaf of$M$isomorphic to$\mathbb Z/p$over the above étale cover (in fact it is the twist of$\mathbb Z/p$by exactly this cover). This extends without problem to twisted actions of the formal multiplicative group. Such thingies do occur in practice. My favourite example is the following: Look at the moduli space (i.e., stack if you want to quibble) of ordinary elliptic curves. Formally at each point this space can be identified using Serre-Tate coordinates (which I guess in this$1$-dimensional case is earlier than Serre-Tate) with the formal multiplicative group. We can now consider the (pro)-étale covering (with structure group$\mathbb Z_p^\ast$) trivialising the group of$p$-torsion points of the universal elliptic curve. By Igusa this is connected and on it we have an action of the formal multiplicative group realising formally at each point the Serre-Tate coordinates. 3 added 16 characters in body This is not an answer to the questions but some general comments. One should be aware that the relation between vector fields and Hasse derivations in characteristic$p$is not at all analogous to the characteristic$0$. It is true that a Hasse derivation in all characteristics is the same thing as an acction of the formal additive group. The difference is whereas in characteristic$0$the formal additive group is the only$1$-dimensional formal group whereas in positive characteristic there are more, for instance the formal multiplicative group. In positive characteristic the derivation part of a Hasse derivation corresponds to an action of the group scheme that is the kernel of the Frobenius map which is$\alpha_p$. Again there are several group schemes of order$p$(such as$\mu_p$which is the kernel of Frobenius on the formal multiplicative group). On the vector field side there thus is a first obstruction on the vector field itself, that it should give rise to an$\alpha_p$-action. Concretely that means that the$p$-th power of the derivation should be$0$. Note that that means that there may not even be a vector field to start with even though there might be many vector fields on$M$there may not be any with that property (for instance a smooth proper toric variety with no automorphisms outside of the torus). In any case if one wants an obstruction theory one should note that given$D_1$,$D_n$for$n=1,\ldots,p-1$is just$D_1^n$so the first undetermined one would be$D_p$. If one has local liftings one should compare two such liftings$D_p$and$D'_p$and it follows that their difference$D'_p-D_p$is a derivation and one gets a torsor of the tangent sheaf as a first obstruction. Unfortonately if$D$is a derivation$D'_p=D_p+D$may not fulfil the further condition for being a part of a Hasse derivation namely that its$p$'th power should be zero. One can expand its$p$'th power using the Jacobson formula but it leads to a (seemingly) nasty non-linear problem. Anyway if solvable one can continuer with$D_{2p}$D_{p^2}$ which would be the next undetermined term and continue in the same manner but it looks like a nightmare (the fact that I don't think that local liftings may exist makes it even less palatable).

However, it is clear that you do not have unique extensions: Take some $M$ on which an action of $\hat G_a\times \hat G_a$ is given. Then the actions of $\hat G_a$ given by the inclusions $t\mapsto (t,0)$ and $t\mapsto (t,t^p)$ have the same first order action.

Addendum: I think that all in all Hasse-Schmidt derivations are better than Hasse derivations (recall that a Hasse-Schmidt derivation is just a map $D_\infty\times M\to M$, i.e., not necessarily a formal group action). There the liftings of an order $n$ HS-derivation to order $n+1$ is just a pseudo-torsor over the tangent sheaf with local liftings existing so that the obstruction for extension is just an $H^1$. Of course you will have many more of them but as you don't have uniqueness for lifting vector fields to Hasse derivations anyway it seems to matter less.