1. There are certainly many cases where $f^*L$ is still very ample: e.g. if $X$ is an irreducible curve, any line bundle on $X$ of degree $\geq 2g(\tilde{X} )+1$ will have this property.

2. A typical situation where $f^*L$ is not very ample is when $H^0(\tilde{X},f^*L )\cong H^0(X,L)$. For instance, take for $X$ a curve with one ordinary double point $s$, and $L=\omega _X(-p)$, $p$ a smooth point of $X$. Then $f^*L=\omega _{\tilde{X}}(s'+s''-\tilde{r} )$, where $p,q$ are the 2 points of $\tilde{X} $ above $s$ and $\tilde{r} $ the point above $r$. Riemann-Roch gives $h^0(f^*L)= g(\tilde{X} )=h^0(L)$ (I choose $p$ so that $h^0(s'+s''-\tilde{p} )=0$).

I think there is no simple criterion allowing to decide which situation occurs, you have to work.

1. There are certainly many cases where $f^*L$ is still very ample: e.g. if $X$ is an irreducible curve, any line bundle on $X$ of degree $\geq 2g(\tilde{X} )+1$ will have this property.

2. A typical situation where $f^*L$ is not very ample is when $H^0(\tilde{X},f^*L )\cong H^0(X,L)$. For instance, take for $X$ a curve with one ordinary double point $s$, and $L=\omega _X(-p)$, $p$ a smooth point of $X$. Then $f^*L=\omega _{\tilde{X}}(s'+s''-\tilde{p} )$, where $s',s''$ are the 2 points of $\tilde{X} $ above $s$ and $\tilde{p} $ the point above $p$. Riemann-Roch gives $h^0(f^*L)= g(\tilde{X} )=h^0(L)$ (I choose $p$ so that $h^0(s'+s''-\tilde{p} )=0$).

I think there is no simple criterion allowing to decide which situation occurs, you have to work.