To answer your question I'm going to prove something a little more general: I'm going to consider a generic map $f \colon \mathbb R^{m+1} \to \mathbb R^{n+1}$.
The well definition problem for the function is that for every $x_1,x_2 \in X$ such that $[x_1]=[x_2]$ the equality $[f(x_1)]=[f(x_2)]$ holds.
Let $\pi_m \colon \mathbb R^{m+1}\setminus\{0\} \to \mathbb{RP}^m$ be the canonical projection, i.e. the map sending every $x \in \mathbb R^{m+1}\setminus\{0\}$ in $\pi_m(x)=[x] \in \mathbb{RP}^m$. Now the well definition problem can be rephrased in terms of diagrams saying that we want find necessary and sufficient conditions for the commutativity of the diagram below .

This function $[f]$ exists if and only if for each pair $x,y \in \mathbb R^{m+1}\setminus\{0\}$ $[x]=[y]$, i.e such that exists $\lambda \in \mathbb R \setminus \{0\}$ the condition $[f(x)]=\pi_n \circ f(x)=\pi_n \circ f(y)=[f(y)]$, this by universal property of quotient set for a given equivalence relation.
Edit: I'm adding some details about the said universal property for quotient sets.
Let $\pi \colon X \to Y$ be a surjective function. Then for every other function $g \colon X \to Z$ there exists a (necessarily unique) $h \colon Y \to Z$ such that
$ g = h \circ \pi $ (i.e. $f$ factors through $\pi$) if and only if for each $x_1,x_2 \in X$ such that $\pi(x_1)=\pi(x_2)$, we have that $g(x_1)=g(x_2)$.
Proof: If this $h$ exists clearly we have that for each pair $x_1,x_2 \in X$ such that $\pi(x_1)=\pi(x_2)$, we have that
$g(x_1)=h \circ \pi(x_1)=h \circ \pi(x_2) = g(x_2)$
If $k \colon Y \to Z$ is another function such that $k \circ \pi=g$ then we have the for every $y \in Y$ exists a $x \in X$ such that $\pi(x)=y$, and so
$h(y)=h \circ \pi(x) = k \circ \pi(x)=k(y)$
and so $k=h$ (thus if $h$ exists it's unique too).
Let's now show that if the condition holds then function $h$ as above exists. For each $x \in X$ we can consider the set $\left\{g(x') | x' \in X, g(x')=g(x) \right\}$, by the hypothesis this set contains just one element, namely $g(x)$. So its well defined tha map $h(\pi(x))=g(x)$, because $g(x)$ doesn't depend of the choice of the $x \in X$. Clearly this $h$ is the map we were looking for.
Now this theorem can be applied to the problem in the begging letting be $X= \mathbb R^{m+1} \setminus \{0\}$, $Y=\mathbb{RP}^{m}$, $\pi=\pi_m$ , $Z=\mathbb{RP}^{n}$ and $g=\pi_n \circ f$.
In particular if $f \mathbb R^{m+1} \to \mathbb R^{n+1}$ is a linear injective map this we have the map $[f]$ because for every $x_1,x_2 \in \mathbb R^{m+1} \setminus \{0\}$ such that exists $\lambda \in \mathbb R \setminus \{0\}$ for which $x_2=\lambda x_1$ we have that $f(x_2) = f(\lambda x_1)=\lambda f(x_1)$ and thus $\pi_n \circ f(x_2) = \pi_n (\lambda f(x_1)) = \pi_n \circ f(x_1)\ \text{.}$
Injectivity is required because otherwise we could have points of $\mathbb R^{m+1} \setminus \{0\}$ which were send to $0$, which doesn't belong to $\mathbb R^{n+1} \setminus \{0\}$ so $[f]$ could not exists in this case.