I'm trying to show that for an ordinary manifold $X$ and a supermanifold $S$, supermanifold morphisms $\varphi:S\to TX$ are one-to-one to the pairs $(f,F) $ where $f:C^\infty(X)\to C^\infty(S)$ is a super $\mathbb{R}$-algebra homomorphism and $F:C^\infty(X)\to C^\infty(S)$ is an even derivation with respect to $f$, i.e. $F$ is a parity-preserving $\mathbb{R}$-linear map and $F(ab)=F(a)f(b)+f(a)F(b)$ for any $a,b\in C^\infty(X)$.

Intuitively, $f$ is the pullback of smooth maps from the base manifold $X$ to $S$ and $F$ is a global section to the pullback vector bundle $f^*TX$ on $S$. Hence if we let $p:TX\to X$ be the usual projection, then $f= \varphi^*\circ p^* $. The remaining data of $\varphi$ should give $F$, but I cannot see how.

If $S$ is also an ordinary manifold, then $F$ can be defined point-wisely. We can express $\varphi$ point-wisely by $\varphi(p)=(f(p),V_{f(p)})$ for each $p\in S$, where $f(p)\in X$ and $V_{f(p)}\in T_{f(p)}X$. For any $a\in C^\infty(X)$, $F(a)\in C^\infty(S)$ is then the function that sends $p\in S$ to $V_{f(p)}a\in \mathbb{R}$.

However, if $S$ is a supermanifold, then the point-wise definition no longer works. I tried to formulate the point-wise definition in a non-point-wise way but failed.

I guess that $F$ should be a composition $C^\infty(X)\xrightarrow{D}C^\infty(TX)\xrightarrow{\varphi^*}C^\infty(S)$ for some natural derivation $D:C^\infty(X)\to C^\infty(TX)$ with respect to $p^*$, but I cannot see any derivation that arises "naturally".

Thanks in advance for any help.