Consider the steady flow of a fluid (that is, the velocity at each point of the flow does not change over time). In time t a particle at point x of the flow will move to some new point $f_{t}(x)$ of space. The mapping $x↦f_t(x)$ that arises thereby on the points of space occupied by the flow depends on time and is called the mapping after time t. Show that $f_{t2}∘f_{t1}=f_{t1}∘f_{t2}=f_{t1}+t_2$ and $f_t∘f_{−t}=e_X$.
My attempt: Let $f_t$ be defined as $f_t(x) = x + vt$. Then $$f_{t2} \circ f_{t1} = f_{t2}(x+vt_1) = (x+vt_1)+vt_2$$ $$f_{t_1} \circ f_{t_2}(x) = f_{t_1}(x+vt_2) = (x+vt_2)+vt_1$$ $$f_{t_1+t_2}(x) = x + v(t_1+t_2) = x + vt_1 +vt_2$$ $$f_t \circ f_{-t}(x) = f_t(x + (-vt)) = (x-vt) + vt = x$$
Does this resemble the truth? Not sure about "$f_t$ being defined as $f_t(x) = x + vt$" being correct in particular, but logically seems sound I think.
You solution is incorrect, since you cannot assume that the velocity $v$ is constant. We can assume we have a velocity field $v(x)$ (which doesn't depend on $t$), and thus
$x'(t) = v(x)$
Is a regular autonomous ODE of the first order, which it's solution is given by
$f_t(x) = x(t) = x(0) + \int_0^t v(x(a))da$
Now we can go on to prove your exercise -
$ x(s) = x(0) + \int_0^s v(x(a))da$
Solving the ODE again with the initial condition $x(s)$ and predicting the solution at time $t$ gives
$ x(t+s) = x(s) + \int_s^t v(x(a))da $
Substituting $x(s)$ from the original formula gives
$ x(t+s) = x(0) + \int_0^s v(x(a))da + \int_s^t v(x(a))da = x(0) + \int_0^{t+s} v(x(a))da $
which is consistent with what you wish to prove - $f_{t+s} = f_t \circ f_s$. The rest of the computations are similar.