The answer to the second question is yes, the image under $f$ of a convex set is convex. In general, the image of a segment $AB$ under $f$ is the segment $f(A)f(B)$ (provided $f$ is well-defined throughout the segment $AB$, which is certainly the case if $A, B \in (0,+\infty)^n$).
Let $H$ be the hyperplane $x_1 + \dots + x_n = 1$. Then $f$ is the central projection onto $H$ with respect to the origin $O$. The domain of $f$ is the complement of the hyperplane $H_0 \colon x_1 + \dots + x_n = 0$, that is, the union of the two open half-spaces determined by $H_0$.
Now assume the segment $AB$ is contained in the domain of $f$, i.e., $A$ and $B$ are on the same side of $H_0$. If $O, A, B$ lie on the same line, there is nothing to prove, as $f$ is constant on the segment $AB$. Otherwise, the whole problem is situated within the plane $\alpha$ through $O, A, B$.
Let $l, l_0$ be the lines of intersection of $\alpha$ with $H, H_0$. Then $A, B$ are points in $\alpha$ on the same side of $l_0$, which we can assume without loss of generality to be the same side where $l$ is. The problem is to see that if $C$ is a point between $A$ and $B$, then $f(C)$ is between $f(A)$ and $f(B)$ on line $l$ (and all points of the segment $f(A)f(B)$ are of this form).
Now this is reduced to a well-known fact of plane geometry: given three rays $r_1$, $r_2$ and $r_3$ all in the same plane, if at least one segment drawn from $r_1$ to $r_2$ intersects $r_3$, then all segments drawn from $r_1$ to $r_2$ intersect $r_3$.
It is enough to apply this fact to the rays $r_1 = OA$, $r_2 = OB$, $r_3 = OC$ and the segments $AB$, $f(A)f(B)$.
To answer the first question, it is enough to prove that if $C$ is a convex subset of $H$, then the portion of $f^{-1}(C)$ lying in a chosen half-space with respect to $H_0$ is also convex. This can be given a similar proof by restricting to a plane as above. Then it is enough to see that the region lying between two rays is convex.
