Since $\Bbb RP^n$ is the quotient of $S^n$ under the group action by $\Bbb Z/2$ identifying antipodes, there's a reasonably easy solution. Because $k[V/G]=k[V]^G$, where the superscript denotes invariants, we can just calculate the invariants of the group action. The group action on $\Bbb R[x_1,\cdots,x_n]/(1-\sum x_i^2)$ is by $x_i\mapsto -x_i$, which has algebra of invariants generated by the degree-2 polynomials. This is clearly a finitely-generated $\Bbb R$-algebra, so choosing generators and computing relations, one can find the variety you request.
In the case $n=2$, you get the coordinate algebra $$\Bbb R[x_1^2,x_1x_2,x_1x_3,x_2^2,x_2x_3,x_3^3]/(x_1^2+x_2^2+x_3^2-1)$$ which can be written as $\Bbb R[y_1,\cdots,y_6]/I$ for $y_1=x_1^2$, $y_2=x_2^2$, $y_3=x_3^2$, $y_4=x_1x_2$, $y_5=x_1x_3$, $y_6=x_2x_3$ where $I$ is generated by $y_1+y_2+y_3-1$ and all the appropriate quadratic relations of the form $y_iy_j-y_ky_l$ like $y_1y_2-y_4^2$. (These are somewhat annoying to write out, please forgive me for not doing all of them.) This is not the minimal embedding, though - for instance, making the substitution $y_3=1-y_1-y_2$, we can get down to 5 variables, showing that $\Bbb RP^2$ can be represented as the real points of some variety in $\Bbb R^5$.
This isn't great for trying to find the minimal dimension, though - we're embedding $\Bbb RP^n$ in $\Bbb R^{\binom{n+1}{2}-1}$. We know that for any smooth affine variety of dimension $n$, there's an embedding in affine space of dimension $2n+1$ - projecting from a point which isn't on any line through our variety with target a hyperplane produces a map which restricts to an isomorphism on our variety but drops the dimension of the ambient space by 1. (This is roughly the analogue of the Whitney embedding theorem from topology.) In the case $n=2$, it happens that we can actually project one further time: this is related to some very special properties of the Veronese surface. Topological properties show we can't do any better: $\Bbb RP^2$ does not embed in $\Bbb R^3$ for orientability reasons.
The link from the comments gives an example of a polynomial map $\Bbb R^3\to\Bbb R^4$ by $(x,y,z)\mapsto (x^2-y^2,xy,xz,yz)$ which when restricted to $S^2\subset\Bbb R^3$ gives a double covering of its image and therefore a polynomial embedding of $\Bbb RP^2$ in to $\Bbb R^4$. You've asked how to calculate the equations for the image. The techniques for this belong to an area called elimination theory. Briefly, what we're looking to do is to compute the inverse image of the ideal $(x^2+y^2+z^2-1)$ under the ring map $\Bbb R[a,b,c,d]\to \Bbb R[x,y,z]$ sending $a\mapsto x^2-y^2$, $b\mapsto xy$, $c\mapsto xz$, $d\mapsto yz$. After a little computer assistance, we find that the ideal in question is $$(a^2d + 4b^2d + c^2d + d^3 - 2bc + ad,\\ bc^2 - acd - bd^2,\\ 2b^2c - abd + cd^2 - bd,\\ abc + 2b^2d + c^2d - bc,\\ a^2c + c^3 + 2abd - cd^2 - ac,\\ a^2b + 4b^3 + acd + 2bd^2 + cd - b,\\ a^3 + 4ab^2 + ac^2 + ad^2 + c^2 - d^2 - a).$$ I hope one additional thing you take from this is that sometimes looking for explicit equations can be less convenient than knowing you have a variety by other means.