Let $M,N$ be $d$-dimensional Riemannian manifolds, $f:M \to N$ a smooth map. Then $df \in \Omega^1\big({M,f^*TN}\big)=\Gamma(T^*M \otimes f^*TN)$.
Let $\nabla$ be the pullback connection on $f^*TN$ induced by the Levi-Civita connection on $TN$.
Does there exist an element $\sigma \in \Gamma(f^*TN)$ such that $\nabla \sigma=df$?
Denote by $d_{\nabla}:\Omega^k\big({M,f^*TN}\big) \to \Omega^{k+1}\big(M,f^*TN\big)$ the associated exterior derivative.
Then, if such a $\sigma$ exist, then $-R^{f^*TN} \wedge \sigma =d_{\nabla} d_{\nabla} \sigma=d_{\nabla} \nabla \sigma=d_{\nabla} df=0$, where the last equality comes from the symmetry of the connection on $TN$.
Thus,
$R^{f^*TN} \wedge \sigma =0$ is a necessary condition for such a $\sigma$. ($R^{f^*TN}$ is the curvature tensor of $\nabla$).