An exercise in Lee's Introduction to Smooth Manifolds asks one to check that $F:\mathbb R\to\mathbb R^2$ given by $F(t)=(\cos t,\sin t)$ relates $X=d/dt\in\mathfrak X(\mathbb R)$ to $Y\in\mathfrak X(\mathbb R^2)$ given by $$ Y=x\frac\partial{\partial y}-y\frac\partial{\partial x} $$
I have answered this question by assuming that we are given $f\in C^\infty(\mathbb R)$ and checking that for $p\in\mathbb R$ we have $$ dF_p(X_p)f=Y_{F(p)}f $$ This all checks out fine, but is it possible to evaluate the relation $dF_p(X_p)=Y_{F(p)}$ directly without first introducing some $f\in C^\infty(\mathbb R)$ that both sides act on?
Sure. Let us identify, as is common, $T_p \mathbb{R}^{n} \cong \mathbb{R}^{n}$, using the basis $\frac{\partial}{\partial x^1}, \ldots, \frac{\partial}{\partial x^n}$. Under this identification, we have $X_p = (1)$, $Y_{F(p)} = (-\sin p, \cos p)$, and the matrix for $dF_p$ is $[-\sin p\ \cos p]$. Hence it can easily be seen that $dF_p(X_p) = Y_{F(p)}$.