In case $X, Y$ are smooth,
$ f^* $
is full if and only if it is full and faithful. This is explained in the introduction of [arXiv:1101.5931][1] (by Canonaco-Orlov-Stellari), which also studies when this implication holds more generally. Thus the pull-back is full and faithful if and only if $Rf_* \mathcal{O}_X = \mathcal{O}_Y$; for example, when $f$ is birational.

Here is a short explanation of their argument for the case of pull-back functors: $f^*$ is full and faithful if and only if $\mathrm{Ext}^i_{Y}(f^* O_{x_1}, f^* O_{x_2})$ is

 1. $\mathbb C$ for $x_1 = x_2$, $i = 0$,
 2. 0 for $i \notin [0, \mathrm{dim} Y]$, and
 3. 0 for $x_1 \neq x_2$.

(This is due to Bondal-Orlov and Bridgeland.)

Since $f^* $ is full, the 2nd and 3rd condition are automatically satisfied. Thus it will be full and faithful if and only if $f^* O_x \neq 0$ for all $x \in X$. 

Pick $x \in X$ such that $f^* O_x$ is non-zero.
Note that $\mathrm{Hom}^{\bullet}_Y(f^* O_x, f^* O_x) = 
\mathrm{Hom}^{\bullet}_X(O_x, f_* f^* O_x)$ is a quotient of 
$\mathrm{Hom}^{\bullet}_X(O_x, O_x)$, and $f_* f^* O_x$ is supported at $x$. From this one can show that $O_x$ is a sheaf, and in fact isomorphic to $O_x$; all we need is that it's Chern character (in cohomology) is non-zero.
But the Chern character of $f_* f^* O_x$ is independent of $x$, so $f^* O_x \neq 0$ for all $x \in X$.

  [1]: http://arxiv.org/abs/1101.5931