Characterization of being a submersion Let $\mathcal{M}$ be a smooth $m$-dimensional manifold and let $F\colon \mathcal{M} \longrightarrow \mathbb{R}^{n}$ be any smooth map, with m $\geq$ n . I want to prove that, if $p \in \mathcal{M}$:
$$ \text{$F$ is a submersion at $p$} \iff \lbrace d_{p}F^{1},\dots,d_{p}F^{n}\rbrace  \text{ are l.i. in $(T_{p}\mathcal{M})^{*}$} $$
Were $d_{p}f := \lambda^{-1}_{f(p)} \circ T_{p}f \colon T_{p}\mathcal{M} \longrightarrow \mathbb{R} \in (T_{p}\mathcal{M})^{*}$, $T_{p}f\colon T_{p}\mathcal{M}\longrightarrow T_{f(p)}\mathcal{M}$ is the tangent map at $p$ and $\lambda_{p}\colon V \longrightarrow T_{p}V$ is the canonical isomorphism between a finite dimensional vector space over $\mathbb{R}$ and its tangent space. We can extend the definition of differential of a smooth map $F\colon \mathcal{M} \longrightarrow V$ by putting $D_{p}F := \lambda^{-1}_{F(p)} \circ T_{p}F \colon T_{p}\mathcal{M} \longrightarrow V$, where $V$ is a finite dimensional vector space over $\mathbb{R}$.
This is trivial if you take $\mathcal{M} = \mathbb{R}^{m}$ since you can identify the tangent map with the differential from ordinary calculus, but I want to formalize the proof as much as possible in the general case using the definitions I gave.
 A: Linear algebra Fact: Let $A:V\rightarrow \mathbb{R}^n$ be a linear transformation of vector spaces, then $A$ is surjective iff $\pi_1A,\pi_2A,...,\pi_nA$ are linearly independent members of $V^*$. (Where $\pi_i:\mathbb{R}^n\rightarrow \mathbb{R}$ is ith projection).
Proof: Equip $V$with any inner product, then there exists $v_1,...,v_n\in V$ such that $\pi_iA(x)=x \circ  v_i$ for all $x\in V , i\in [n]$. $\dim(Im(A))=\dim(V)-dim(Ker(A))=dim(V)-dim(\{v_1,...,v_n\}^{\perp})=dim(span(\{v_1,...,v_n \}))$. 
Thus $A$ is surjective iff $dim(Im A)=n$ iff $dim(span(\{v_1,...,v_n\}))=n$ iff $v_1,...,v_n$ are L.I. iff $\pi_1A,...,\pi_nA$ are L.I. $\square$
Identify the tangent space of any point of $\mathbb{R}^n$ with $\mathbb{R}^n$
Now let's go back to your setting, then for any $v\in T_pM$: $dF^i_p(v)=<v,F^i>=<v,\pi_i \circ f>=<Df|_pv,\pi_i>=\pi_i(Df|_pv)$.
So $dF^i_p=\pi_i\circ Df|_p$. Now apply the previous fact on the linear transformation $Df|_p:T_pM\rightarrow \mathbb{R}^n$ and you will be done.
