Here is another answer, probably from a slight different angle.

In my opinion, the easiest way to see that $\Omega_{\mathbb P^n}^1(2)$ is globally generated (at least over $\mathbb C$) is the following direct computation. 

Let $\mathbb P^n=\mathbb P(\mathbb C^{n+1})$ be the projective space of lines in $\mathbb C^{n+1}$ with coordinates $(Z_0,\dots,Z_n)$ and $U_j$, $j=0,\dots,n$ be the affine open set in $\mathbb P^n$ corresponding to $Z_j\ne 0$. Finally, let $(z_{j,1},\dots,z_{j,n})$ be the corresponding affine coordinates on $U_j$. The (anti)tautological line bundle $\mathcal O(1)$ on $\mathbb P^n$ is then described by the following transition functions:
$$
g_{\alpha\beta}([Z_0:\cdots:Z_n])=\frac{Z_\beta}{Z_\alpha},\quad U_\alpha\cap U_\beta.
$$ 
Thus, at $[Z_0:\cdots:Z_n]\in U_\alpha\cap U_\beta$, the line bundle $\mathcal O(2)$ has transition functions given by $g_{\alpha\beta}^2([Z_0:\cdots:Z_n])=(Z_\beta/Z_\alpha)^2$.

Now, take any non-zero vector $v$ in $\Omega_{\mathbb P^n,x_0}^1(2)$. Without loss of generality (by acting with $PGL(n)$, rotating and rescaling if necessary), you can suppose that $x_0=[1:0:\cdots:0]\in U_0$ and that $v=(dz_{0,1}\otimes\eta_0)(x_0)$, where $\eta_j$ is a local frame for $\mathcal O(2)$ on $U_j$, so that $\eta_\beta=g_{\alpha\beta}^2\eta_\alpha$ on $U_\alpha\cap U_\beta$.

I claim that the section $dz_{0,1}\otimes\eta_0$, *a priori* defined only over $U_0$, is in fact a global holomorphic section. To see this, it suffices to check what happens when passing from $U_0$ to $U_j$, $1\le j\le n$. We have, for $j=1$, $z_{0,1}=1/z_{1,1}$ and, for $j\ge 2$, $z_{0,1}=z_{j,2}/z_{j,1}$. Therefore,
$$
dz_{0,1}=-\frac{dz_{1,1}}{z_{1,1}^2}=\frac{z_{j,1}dz_{j,2}-z_{j,2}dz_{j,1}}{z_{j,1}^2},\quad j\ge 2.
$$ 
On the other hand, 
$$
\eta_{0}=(Z_0/Z_j)^2\eta_j=z_{j,1}^2 \eta_j.
$$
So,
$$
dz_{0,1}\otimes\eta_0=-dz_{1,1}\otimes\eta_1=(z_{j,1}dz_{j,2}-z_{j,2}dz_{j,1})\otimes\eta_j,\quad j\ge 2,
$$
which are actually holomorphic. 

This proves "by hands" the global generation of $\Omega_{\mathbb P^n}^1(2)$. 

Please note, that this global generation is just a reformulation of the elementary fact that the differential of a meromorphic function with a simple pole is at worst a meromorphic $1$-form with a pole of order $2$! 

Turning to your second question, I think that in this generality you propose one cannot say much more than the following.

Any quotient of a trivial vector bundle is globally generated and, conversely, every globally generated vector bundle $E\to X$ of rank $r$ over a $n$-dimensional complex manifold $X$ is isomorphic to the quotient of a trivial vector bundle of rank $\le n+r$ (this is a non-trivial fact). In particular, if you are able to fit your vector bundle $E$ into a sequence $V\to E\to 0$, where $V$ is globally generated, then $E$ is globally generated itself. 

Thus, for instance, if $X\subset\mathbb P^n$ is a smooth submanifold, you have a short exact sequence of vector bundles
$$
0\to T_X\to T_{\mathbb P^n}|_X\to N_{X/\mathbb P^n}\to 0.
$$
Twisting by $\mathcal O(-2)$ and taking duals, you obtain
$$
0\to N_{X/\mathbb P^n}^*(2)\to\Omega_{\mathbb P^n}^1(2)|_X\to\Omega_X^1(2)\to 0.
$$   
By the above criterion, the cotangent bundle of any (embedded) smooth projective manifold twisted by $\mathcal O(2)$ is globally generated.

If you want more sophisticated criterions, then probably you should restrict a little bit more your second question.