If I am solving a positive-dimensional system of polynomials over $\mathbb{R}$, and specifically am searching only for real solutions, how do I know that my solution is complete and there are no other possibilities? For example, consider the system:
$$x_1^2 + x_2^2 = 1$$
I can solve it by letting $x_1 = t \in [-1,1]$, then $x_2 = \pm \sqrt{1-t^2}$ and I'm golden. As far as I know there are no more real solutions that are unique (excluding the one that is the same up to the choice of parameter). But how do I know that? When the system gets more complicated, is there a nice way to figure it out?
I'd be interested in a general proof over an arbitrary field as well. However, I suspect my life is going to be made harder because I am working with the real solutions only, and as a result there are solutions over the field extension that I don't consider.
Specifically, the equations are:
$$f\sum_{i=1}^n x_{i4} - K1 = 0$$
$$f\sum_{i=1}^n x_{i3}x_{i2} - K_2 = 0$$
$$f\sum_{i=1}^n x_{i1}x_{i3} - K_3 = 0$$
$$f\sum_{i=1}^n p_{z1}x_{i2}x_{i3} + f\sum_{i=1}^np_{y1}x_{i1}x_{13} - K_4 = 0$$
$$f\sum_{i=1}^n p_{z1}x_{i4} - f\sum_{i=1}^np_{x1}x_{i1}x_{13} - K_5 = 0$$
$$f\sum_{i=1}^n -p_{y1}x_{i4} - f\sum_{i=1}^np_{z1}x_{i1}x_{13} - K_6 = 0$$
along with the "circularity equations":
$$x_{i1}^2 + x_{12}^2 = 1$$
$$x_{i3}^2 + x_{14}^2 = 1$$
where $f \in \mathbb{R}^+$, $K_j \in \mathbb{R}$ (there is technically another set of circularity equations but they are on variables that do not appear in the above system).
My answer ended up becoming rather much longer than I had hoped, even though I have omitted quite some details. So first I'd like to remark on your original question about general systems of polynomial equations. In the comments I said solving your system is relatively easy because it concerns only polynomials of low degree (in fact it it reduces to a system of equations of degree at most $2$). In spite of my unpleasantly long and cumbersome answer I stand by this comment; in general, solving systems with (many) polynomials of higher degrees is a very ad hoc matter, and is more often than not practically intractible.
In finding all solutions I'll assume familiarity with basic linear algebra; some vector and matrix notation, and solving systems of linear equations. Let ${\bf1}_n=(1,1,\ldots,1)\in\Bbb{R}^n$, let $K_i':=\tfrac{K_i}{f}$ and define $x_1,x_2,x_3,x_4\in\Bbb{R}^n$ as $$x_j:=(x_{ij})_{1\leq i\leq n}.$$ This allows us to write down the system of six equations more concisely and transparently. Rewriting a bit and substituting the first two equations into the last three, the system becomes \begin{eqnarray*} \langle x_4,{\bf1}\rangle&=&K_1',\\ \langle x_2,x_3\rangle&=&K_2',\\ \langle x_1,x_3\rangle&=&K_3',\\ p_{z1}K_2'+p_{y1}x_{13}\langle x_1,{\bf1}_n\rangle&=&K_4',\\ p_{z1}K_1'+p_{x1}x_{13}\langle x_1,{\bf1}_n\rangle&=&K_5',\\ p_{y1}K_1'+p_{z1}x_{13}\langle x_1,{\bf1}_n\rangle&=&-K_6'. \end{eqnarray*} What is most striking is that, apart from the coefficient $x_{13}$ in the last three equations, these equations concern the $x_j$ only in terms of inner products! Also note that the variables $x_{24},\ldots,x_{n4}$ appear only in the first equation, which is linear and hence easy to solve. So will omit the first equation from now on.
First we treat some boundary cases. If $x_{14}=\pm1$ then $x_3=0$ and the system of equations becomes \begin{eqnarray*} 0&=&K_2',\\ 0&=&K_3',\\ 0&=&K_4',\\ p_{z1}K_1'&=&K_5',\\ p_{y1}K_1'&=&-K_6'. \end{eqnarray*} This is a linear system of equations, having solutions (with $x_{14}=\pm1$) if and only if if $K_i'=0$ for all $i$, or $K_2'=K_3'=K_4'=0$ and $K_1'\neq0$. Similarly, if $x_{12}=\pm1$ then $x_1=0$ and the system of equations becomes \begin{eqnarray*} \langle x_2,x_3\rangle&=&K_2',\\ 0&=&K_3',\\ p_{z1}K_2'&=&K_4',\\ p_{z1}K_1'&=&K_5',\\ p_{y1}K_1'&=&-K_6'. \end{eqnarray*} For any choice of $x_{22},\ldots,x_{n2}\in\Bbb{R}$ this is again a linear system. There are no solutions (with $x_{12}=\pm1$) if and only if $K_1'K_4'\neq K_2'K_5'$, or $K_2'=0$ and $K_4'\neq0$, or $K_1'=0$ and $K_5'\neq0$ or $K_6'\neq0$.
Now let $x_{12},x_{14}\in(-1,1)$. It follows from the circularity equations that for all $i$ $$x_{i1}=\pm\sqrt{1-x_{12}^2}\qquad\text{ and }\qquad x_{i3}=\pm\sqrt{1-x_{14}^2}.$$ Hence $x_{i1}=\pm x_{11}$ and $x_{i3}=\pm x_{13}$ for all $i$, so there exist unique vectors $e_1,e_3\in\Bbb{R}^n$ satisfying $$x_1=\sqrt{1-x_{12}^2}\ e_1\qquad\text{ and }\qquad x_3=\sqrt{1-x_{14}^2}\ e_3,$$ and all their coefficients are either $1$ or $-1$. We may now rewrite the equations as \begin{eqnarray*} \sqrt{1-x_{14}^2}\langle x_2,e_3\rangle&=&K_2',\\ \sqrt{1-x_{14}^2}\sqrt{1-x_{12}^2}\langle e_1,e_3\rangle&=&K_3',\\ p_{z1}K_2'+p_{y1}e_{13}\sqrt{1-x_{12}^2}\sqrt{1-x_{14}^2}\langle e_1,{\bf1}_n\rangle&=&K_4',\\ p_{z1}K_1'+p_{x1}e_{13}\sqrt{1-x_{12}^2}\sqrt{1-x_{14}^2}\langle e_1,{\bf1}_n\rangle&=&K_5',\\ p_{y1}K_1'+p_{z1}e_{13}\sqrt{1-x_{12}^2}\sqrt{1-x_{14}^2}\langle e_1,{\bf1}_n\rangle&=&-K_6'. \end{eqnarray*} This eliminates $x_1$ and $x_3$ and leaves only $e_1$, $e_3$ in stead in these equations. First consider the case that $m:=\langle e_1,e_3\rangle\neq0$. Then by the second equation we have $K_3'\neq0$, and it can be written as $$\sqrt{1-x_{14}^2}\sqrt{1-x_{12}^2}=\frac{K_3'}{m},$$ which shows that we must have $\tfrac{K_3'}{m}\in[-1,1]$, or equivalently $|K_3'|<|m|$. In particular we must have $|K_3'|<n$ for solutions to exist at all, as $|m|\leq n$. So let's assume that $|K_3'|<n$, and choose $e_1$ and $e_3$ such that $|K_3'|<|m|$. Counting the number of pairs for which this holds as a function of $K_3'$ is a nice combinatorial problem that is suited for a question of it's own, though it is not complicated.
The first equation is now a linear equation for $x_{22},\ldots,x_{n2}\in\Bbb{R}$. Substituting the second equation into the last three yields the system \begin{eqnarray*} \sqrt{1-x_{14}^2}\sqrt{1-x_{12}^2}\langle e_1,e_3\rangle&=&K_3',\\ p_{z1}K_2'+p_{y1}e_{13}\frac{K_3'}{\langle e_1,e_3\rangle}\langle e_1,{\bf1}_n\rangle&=&K_4',\\ p_{z1}K_1'+p_{x1}e_{13}\frac{K_3'}{\langle e_1,e_3\rangle}\langle e_1,{\bf1}_n\rangle&=&K_5',\\ p_{y1}K_1'+p_{z1}e_{13}\frac{K_3'}{\langle e_1,e_3\rangle}\langle e_1,{\bf1}_n\rangle&=&-K_6', \end{eqnarray*} which are then also linear equations in $p_{x1},p_{y1},p_{z1}\in\Bbb{R}$, where there are unfortunately a lot of cases to distinguis. The first equation simply tells us that $$x_{14}^2=1-\frac{K_3'^2}{(1-x_{12}^2)\langle e_1,e_3\rangle^2},$$ which determines $x_{14}$ up to sign.
Finally, consider the case that $\langle e_1,e_3\rangle=0$. Note that this is only possible if $n$ is even. Then $K_3'=0$ and the second equation is obsolete. For each of the $2^n$ choices for $e_1$ there are $\tbinom{n}{\tfrac{n}{2}}$ choices for $e_3$, by changing the signs of precisely half of all coordinates of $e_1$. For every choice of $x_{12},x_{14}\in\Bbb{R}$ the remaining equations are again linear, and there are a lot of cases to distinguish as before.