Definition of derivatives on vector fields on manifolds

While studying the definition of related vector fields for my course in differentiable manifolds, I noticed the following:

We gave the following propositions about the derivative of a function $$f: M \to N$$:

1. We have $$d_pf: T_pM \to T_{f(p)}N$$

2. Taking $$X_p \in T_pM$$ as a differential operator, we have that $$(d_pfX_p)(g)=X_p(g \circ f)$$ for $$g \in C^{\infty}(N)$$

How can these two definitions be compatible if $$X_p$$ as a differential operator is $$X_p: C^{\infty}(M) \to C^{\infty}(M)$$?

The only way I see for the result of $$(d_pfX_p)(g)$$ to be in $$T_{f(p)}N$$ is if $$(d_pfX_p)(g)= Y_{f(p)}(g)$$, were $$Y\in T_{f(p)}N$$. That means that there is a vector field Y f-related to X. But, as far as I understand, there is no guarantee that there is such a vector field Y. So this way of making the two propositions compatible seems wrong.

I feel I am missing something important.

$$X_p\in T_pM$$, is a map $$C^{\infty}(M)\to\mathbb{R}$$, so for $$h:M\to\mathbb{R}$$ we have $$X_ph\in\mathbb{R}$$
$$d_pf.X_p\in T_{f(p)}N$$, is a map $$C^{\infty}(N)\to\mathbb{R}$$, so for $$g:N\to\mathbb{R}$$ we have $$(d_pf.X_p)g=X_p(g\circ f)\in\mathbb{R}$$
So indeed $$d_pf$$ is a map $$T_pM\to T_{f(p)}N$$
You might be confusing the tangent vector $$X_p\in T_pM$$ with some section $$Z\in\Gamma(TM)$$, i.e., a map $$M\to TM$$ such that $$\pi\circ Y=\mathrm{id}_M$$ (where $$\pi$$ is the bundle projection) and hence $$Z$$ can be seen as a map $$C^{\infty}(M)\to C^{\infty}(M)$$ that acts like $$h\mapsto Zh$$ where $$(Zh)_p:=Z_ph$$.