If $f: S_1\to S_2$ is a smooth map betweentwo regular surface $S_1$ and $S_2$, then $d_pf :T_p(S_1)\to T_{f(p)}(S_2)$ is a linear map for any $p\in S_1$
I cannot prove this statement.
I think that I need to use these followings; $$d_pf(X_p+Y_p)=(d_pf)(X_p)+(d_pf)(Y_p)$$ and $$(d_pf)(cX_p)=c(d_pf)(X_p)$$
Given a smooth map $F: X \rightarrow Y$ between manifolds, the pushforward map $DF: T_pX \rightarrow T_{F(p)}Y$ is defined ( in terms of vector fields, as in your question; as Neal said, this can also be defined in terms of curves. ) by , given a function $g: Y \rightarrow \mathbb R$: $$DF_p(X)(g):=X_p(goF) $$ ,where $X_p$ acts like a directional derivative, using the chain rule on the composition $goF$ that is, the derivation/tangent vector $X_p$ based initially at $p$ in $X$ , is sent to the derivation $X_p$ based at $F(p)$ , where it acts on the composition $goF$. Let $Y_p$ be another tangent vector based at $p$. We want to show: $$DF_p(X_p+Y_p)=DF_p(X_p)+DF_p(Y_p) $$ . But, what is $DF_p(X_p+Y_p)(g)$ ? Using the definitions, this is just : $$(X_p+Y_p)(g) $$. Now, $X_p$ and $Y_p$ are, or can be seen as, directional derivatives, and directional derivatives act linearly, so that $$ (X_p+Y_p)(g)= X_p(g)+Y_p(g)= DF_p(X_p)+DF_p(Y_p)$. A similar argument works for the second part. Can you do it?