Dear CX, Taking a short break from revising: I think this fails in positive characteristic. Let $k$ be a field of characteristic $p>0$. Let $m>1$ be an integer.
Let $\mathbb{A}^6_k$ have coordinates $x_0,x_1,x_2,y_0,y_1,y_2$. Consider the affine hypersurface $X$ with defining equation $f = x_0^py_0+x_1^py_1 - x_2^{pm}y_2$. By the Jacobian criterion, the singular locus is contained in the closed subset $Z(x_0^p,x_1^p,x_2^{pm})$, which has codimension $2$ in $X$. Since $X$ is Cohen-Macaulay and regular in codimension $1$, $X$ is normal by Serre's criterion. Let $R$ denote the coordinate ring of $X$, i.e., $R=k[x_i,y_j]/\langle f \rangle$. By direct computation, the sheaf of relative differentials is the $R$-module with presentation
$\Omega_{R/k} = R\{dx_i,dy_j\}/\langle x_0^pdy_0 + x_1^pdy_1 - x_2^{pm}dy_2\rangle$. In particular, consider the derivation $$\theta:\Omega_{R/k} \to R, \ dy_j \mapsto 0, dx_0 \mapsto 1, dx_1 \mapsto 0, dx_2\mapsto 0.$$ Consider the (partial) resolution $\nu:\tilde{X} \to X$ where $\tilde{X}$ is the affine $5$-space with coordinates $u_0,u_1,x_2,y_0,y_1$, and where $\nu(u_0,u_1,x_2,y_0,y_1)$ is $$ (x_0,x_1,x_2,y_0,y_1,y_2) = (u_0x_2^m,u_1x_2^m,x_2,y_0,y_1, u_0^py_0+u_1^py_1).$$ The exceptional set is the zero set of $x_2$. On the complement of this open set, $\theta$ is the derivation $(1/x_2)^m \partial/\partial u_0$. This derivation has a pole of order $m$ along the exceptional set. Thus, this derivation does not have log poles on the exceptional set.
$\textbf{Edit.}$ Changed pole order $2$ to arbitrary pole order $m$.
