Natural map $f^{\ast}f_\ast\mathcal{L}\to\mathcal{L}$ Let $f\colon X\to Y$ be a morphism of smooth varieties and $\mathcal{L}$ an invertible sheaf on $X$. 
How is the "natural map" 
$$f^{\ast}f_\ast\mathcal{L}\to\mathcal{L}$$
defined (which should be surjective if $f$ is affine) ?
 A: $f^*$ and $f_*$ are adjoint functors, i.e. there is a functorial bijection
$$\operatorname{Hom}(f^* \mathcal G, \mathcal F) = \operatorname{Hom}(\mathcal G, f_*\mathcal F).$$
In particular, for any coherent sheaf $\mathcal F$ on $X$ we have
$$\operatorname{Hom}(f^* f_*\mathcal F, \mathcal F) = \operatorname{Hom}(f_*\mathcal F, f_*\mathcal F),$$
thus the identity map on $f_*\mathcal F$ gives rise to a natural map
$$f^*f_*\mathcal F \to \mathcal F$$
In the affine case $\operatorname{Spec} B \to \operatorname{Spec} A$ with $\mathcal F = M^\sim$, this map is given by
$$M \otimes_A B \to M, m \otimes b \mapsto bm,$$
which is cleary surjective.
In general we cannot reduce to the affine case, but we can reduce to the affine case if $f$ is affine, since in this case around any point $x \in X$ we find some affine open $U \subset Y$, such that $f^{-1}(U) \ni x$ is open affine in $X$. We can check the surjectivity now locally on $f^{-1}(U)$, hence we are in the affine case, where surjectivity holds.
Note that we do not need $\mathcal F$ to be invertible or locally free. We also do not need $X,Y$ to be smooth.

Let me also give you an example, where the map is not surjective. Let $X \xrightarrow{f} \operatorname{Spec} k$ be some projective morphism (i.e. $X$ is a projective $k$-variety) and $\mathcal L$ a line bundle without global sections, for example $\mathcal L = \mathcal O_X(-1)$.
On $\operatorname{Speck} k$ - via $H^0(\operatorname{Spec} k,-)$ - a sheaf is the same as a $k$-vector space, i.e. we have
$$f_*\mathcal L=H^0(\operatorname{Spec} k,f_* \mathcal L) = H^0(X,\mathcal L)=0.$$
Thus $f^*f_*\mathcal L \to \mathcal L$ will not be surjective.
In general, for $f: X \to \operatorname{Spec} k$, the map $f^*f_* \mathcal F \to \mathcal F$ is surjective if and only if $\mathcal F$ is globally generated, since it is the map
$$H^0(X, \mathcal F) \otimes_k \mathcal O_X \to \mathcal F.$$
