Since you write ${\rm Diff}_+(M)$ you are probably assuming $M$ is orientable and diffeomorphisms of $M$ are orientation-preserving. Every diffeomorphism of $M$ can be isotoped to take fibers to fibers. This is proved using the assumption that the base surface $F$ has genus at least 2, so $M$ contains vertical incompressible tori (unions of fibers) and every incompressible torus can be isotoped to be vertical. If a diffeomorphism that takes fibers to fibers preserves orientations of the fibers then it is isotopic to a product of twists along vertical tori. These are generated by lifts of Dehn twists in $F$ and twists taking each fiber to itself, i.e., bundle automorphisms.

The other possibility is a diffeomorphism taking fibers to fibers but reversing their orientations, hence also reversing orientation of $F$ since the orientation of $M$ is preserved. Such diffeomorphisms certainly exist for the product bundle $M=F\times S^1$, and they also exist for any nontrivial bundle whose Euler class is even. Namely, start with an example for $F\times S^1$ and modify this by removing two vertical solid tori $V_1$ and $V_2$ that are interchanged by the diffeomorphism, then glue $V_1$ and $V_2$ back in by diffeomorphisms of their boundary tori preserving fibers and taking a slope $0$ curve to a slope $n$ curve, using the same $n$ in both cases. This gives a circle bundle with Euler class $2n$. The diffeomorphism of $M-(V_1\cup V_2)$ reversing orientations in fiber and base extends over the reglued solid tori since slopes in a torus $S^1\times S^1$ are preserved by a $180$ degree rotation that reflects each $S^1$ factor.

I haven't thought about whether one can do an analogous construction when the Euler class is odd.

Since you write ${\rm Diff}_+(M)$ you are probably assuming $M$ is orientable and diffeomorphisms of $M$ are orientation-preserving. Every diffeomorphism of $M$ can be isotoped to take fibers to fibers. This is proved using the assumption that the base surface $F$ has genus at least 2, so $M$ contains vertical incompressible tori (unions of fibers) and every incompressible torus can be isotoped to be vertical. If a diffeomorphism that takes fibers to fibers preserves orientations of the fibers then it is isotopic to a product of twists along vertical tori. These are generated by lifts of Dehn twists in $F$ and twists taking each fiber to itself, i.e., bundle automorphisms.

The other possibility is a diffeomorphism taking fibers to fibers but reversing their orientations, hence also reversing orientation of $F$ since the orientation of $M$ is preserved. Such diffeomorphisms certainly exist for the product bundle $M=F\times S^1$, and they also exist for any nontrivial bundle whose Euler class is even. Namely, start with an example for $F\times S^1$ and modify this by removing two vertical solid tori $V_1$ and $V_2$ that are interchanged by the diffeomorphism, then glue $V_1$ and $V_2$ back in by diffeomorphisms of their boundary tori preserving fibers and taking a slope $0$ curve to a slope $n$ curve, using the same $n$ in both cases. This gives a circle bundle with Euler class $2n$. The diffeomorphism of $M-(V_1\cup V_2)$ reversing orientations in fiber and base extends over the reglued solid tori since slopes in a torus $S^1\times S^1$ are preserved by a $180$ degree rotation that reflects each $S^1$ factor.

Added a few minutes later: It looks like a similar construction can be made more simply using one vertical solid torus instead of two, where this solid torus projects to a disk neighborhood of a fixed point of an orientation-reversing diffeomorphism of $F$. In this case there is no need to assume the Euler class is even.