Just to complete Georges's answer: let $S$ be an effective divisor on $X$ (i.e. a finite sum of closed points), the Riemann-Roch formula (with Serre duality) is $ h^0(Ω_X(−S))=h^0(O_X(S))+g-1-deg S\ge 1+g-1-\deg S=g-\deg S.$ So the answer to your third question is positive when $\deg S\le g-1$ (this includes the case $S=P$ and $g=2$).
Things become more interesting when $\deg S=g$. The above arguments show that $h^0(\Omega_X(-S))> 0$ if and only if $h^0(O_X(S))\ge 2$. One can identifiy the set of effective divisors of degree $g$ on $X$ to the symetric product $X^{(g)}$ (which is $X^g$ quotient by the symetric group in $g$ elements acting on $X^g$ by permutation of the coordinates). Fix a point $x_0\in X$, then we have a morphism to the Jacobian of $X$ $ f: X^{(g)} \to J(X), \quad S \mapsto [S-gx_0].$ It is well known that $f$ is birational. But what is the exceptional locus of $f$ ? It consists in those $S$ such that $\dim f^{-1}(f(S))>0$. As $f^{-1}(f(S))=|S|$ the linear system of effective divisors linearly equivalent to $S$ and has dimension (as projective variety) $h^0(O_X(S))-1$, we see that
when $\deg S=g$, $h^0(\Omega_X(-S))> 0$ if and only if $S$ belongs to the exceptional locus of $f$.
Edit Answer to Question 2. Let $t$ be a rational function which is an uniformizing element of $X$ at $P$, let $f=t^2$. Then $df=2tdt$ has a simple zero at $P$.
Remark on the divisor of a rational section (to respond to a question raised in the comments). Let $L$ be an invertible sheaf on $X$, let $s$ be a non-zero rational section of $L$ (i.e. $s\in L(U)$ for some non-empty open subset $U$ of $X$ and $s\ne 0$). Then the divisor $\mathrm{div}_L(s)$ is a Cartier divisor such that $O_X(\mathrm{div}_L(s))\simeq L$.
This is can be check by writing $L$ on an open covering $\{ X_i\}_i$ of $X$ as $L|_{X_i}=e_i O_{X_i}$. So $s=e_if_i$ for some $f_i\in K(X)$. The Cartier divisor $\mathrm{div}_L(s)$ is represented by $\{ (X_i, 1/f_i)\}_i$ and we have $sO_X(\mathrm{div}_L(s))=L$.
In the particular case when $L=\Omega_X$, for any rational differential form $s$, we have $\mathrm{div}_L(s)\simeq K_X$ for any canonical divisor $K_X$ on $X$.