The following is really only a sketch. Feel free to ask for more details.
The coefficients of a polynomial $f$ are equal to elementary symmetric polynomials in the roots of $f$. Since the resultant is a polynomial function in the coefficients of two polynomials $f$ and $g$, it is a symmetric polynomial function in the roots of $f$ and $g$. The definition of the resultant is made in such a way that $\operatorname{res}(f,g)=0$ if (and only if) $f$ and $g$ share a common root. Now, we use the following:
Lemma. Let $R$ be an integral domain and denote by $K$ the algebraic closure of its quotient field. Let $p\in R[X,Y]$ a polynomial such that $p(a,a)=0$ for all $a\in K$. Then, $(Y-X)$ divides $p$.
Proof. Write $p\in R[X][Y]$ as a polynomial in $Y$, i.e. $p=\sum_{i=0}^n p_i Y^i$ with certain $p_i\in R[X]$. In this integral domain, perform division with remainder of $p$ by $Y-X$ to obtain $p=q(Y-X)+r$ for $q\in R[Y,X]$ and $r\in R[X]\subseteq R[X,Y]$. Since
$r(a)=r(a,a)=p(a,a)=0$ for all $a\in K$, we must have $r=0$. Indeed, $K$ is an infinite field because it is algebraically closed and any nonzero polynomial has only finitely many roots. Consequently, $p=(X-Y)\cdot q$ as required.
Applying this lemma to the resultant as a polynomial in the zeros of $f$ and $g$, you get the desired statement.
Let's write down the claims:
A) The line $l=\overline{pq}$ meets $X$.
B) Every pair of homogeneous $F,G\in I(X)$ has a common zero on $l$
C) $Res_{x_n}(F,G)$ vanishes at $q$ for all homogeneous pairs $F,G\in I(X)$.
You state that you understand that C) is equivalent to an intermediate result but aren't sure about why C) should be equivalent to A). We'll go through the steps carefully.
First, we'll establish that A) and B) are equivalent (you didn't mention an issue with this, but I want to make sure we cover it anyways, plus it's short).
A) clearly implies B): any $F,G\in I(X)$ will have a common zero at every point in $l\cap X$ which is assumed nonempty.
For the other direction of the equivalence, we prove the contrapositive: if $X\cap l=\emptyset$, then there exist homogeneous $F,G\in I(X)$ so that $F,G$ have no common zero on $l$. Assume $X\cap l=\emptyset$. Up to a change of coordinates, we may assume $l=V(x_2,\cdots,x_n)\subset \Bbb P^n$. Now on $U_0=D(x_0)$, we have $X_0:= X\cap U_0$ and $l_0:= l\cap U_0$ are affine varieties which do not meet. Thus $I_{U_0}(X_0)+I_{U_0}(l_0)=(1)$, so we can find elements $a\in I_{U_0}(X_0)$ and $b\in I_{U_0}(l_0)$ so that $a+b=1$. Then $a$ vanishes on $X_0$ but not $l_0$, and after homogenizing it to $\widetilde{a}$ and multiplying by some power of $x_0$, we get that $x_0^p\widetilde{a}$ is a homogeneous element of $I(X)$ which vanishes only at $[0:1:0:\cdots:0]\in l$. Repeating this construction on $U_1=D(x_1)$, we get a homogeneous element of $I(X)$ which vanishes only at $[1:0:\cdots:0]\in l$. These two elements are our $F,G$ which do not share a common zero on $l$, so B implies A by contrapositive.
Before we start tackling the equivalence of B) and C), let's recall some facts about the resultant:
1) The resultant of two polynomials with coefficients from an integral domain is zero iff they have a common divisor of positive degree.
2) If $A,B$ are two polynomials in $R[x]$ and $\varphi: R\to S$ is a ring homomorphism which extends to a ring homomorphism $\varphi:R[x]\to S[x]$ in the natural way, then:
$Res_x(\varphi(A),\varphi(B))=\varphi(Res_x(A,B))$ if $\deg_x A = \deg_x \varphi(A)$ and $\deg_x B = \deg_x \varphi(B)$
$\varphi(Res_x(A,B))=0$ if $\deg_x A > \deg_x \varphi(A)$ and $\deg_x B > \deg_x \varphi(B)$
$\varphi(Res_x(A,B))=\varphi(a)^{\deg B-\deg \varphi(B)}Res_x(\varphi(A),\varphi(B))$ if $\deg A =\deg \varphi(A)$ and $\deg B > \deg \varphi(B)$ where $a$ is the top coefficient of $A$.
$\varphi(Res_x(A,B))=\pm \varphi(b)^{\deg A-\deg \varphi(A)}Res_x(\varphi(A),\varphi(B))$ if $\deg B =\deg \varphi(B)$ and $\deg A > \deg \varphi(A)$ where $b$ is the top coefficient of $B$.
Each of these parts of 2) can be proven by noticing that $\varphi$ commutes with $\det$ since it's a polynomial.
(See the Wikipedia page if you care about when the $\pm$ is a $+$ versus a $-$.)
Now let's look at the equivalence of B) and C). We'll quantify it as follows: for any pair $F,G\in I(X)$, their having a common zero on $l$ is equivalent to $Res_{x_n}(F,G)(q)=0$.
Suppose either $F$ or $G$ satisfies the condition that it's leading coefficient as a polynomial in $x_n$ doesn't vanish upon evaluation at $q$ (aka restriction to $l$). We apply the fact 2) about the resultant with $\varphi$ being the evaluation at $q$ map: the first, third, or fourth part of this fact applies, and we have that $Res_{x_n}(F,G)(q)=0$ iff $Res_{x_n}(F[q],G[q])=0$. But $Res_{x_n}(F[q],G[q]) = 0$ iff $F[q]$ and $G[q]$ have a common factor of positive degree by fact 1) about resultants, and this common factor is exactly equivalent to a common zero on $l$, so we see that B) and C) are equivalent in this case.
In the case where $F$ and $G$ both have leading coefficients as polynomials in $x_n$ which vanish upon plugging in $q$, we show that conditions B) and C) are automatically true. As $p\Rightarrow q$ is equivalent to $\neg p \vee q$, this will show that B) and C) are equivalent in this case.
If $F,G$ both have leading coefficients as polynomials in $x_n$ which vanish upon plugging in $q$, we are in the situation of the second part of fact 2), so $Res_{x_n}(F,G)(q)=0$. Similarly, the vanishing of the leading coefficient implies that $F,G$ both have a zero at $p$ because $\deg_{x_n} F < \deg F$. (To prove this last bit, it may be instructive to note that up to a change of coordinates leaving $p$ fixed, we may take $q=[1:0:\ldots:0]$, so that $F[q],G[q]$ are either zero or divisible by $x_0$ and therefore must have a zero at $p$.)
I have to admit that I personally got a little turned around a few times attempting to write the last part of this answer - the key thing to note is that there's a case where the equivalence of B) and C) is automatic because they're both just true from the assumptions in this special case. Hope this helps!
Best Answer
Expand $f$ and $g$ as polynomials in $y$ as follows:
$$f= a_0y^d+a_1y^{d-1}+\cdots+a_d$$ $$g= b_0y^e+b_1y^{e-1}+\cdots+b_e.$$
Then the resultant of these two polynomials is given by the determinant of the Sylvester matrix. Here's an example when $f$ has degree three in $y$ and $g$ has degree two in $y$:
$$S= \begin{pmatrix} a_0 & & b_0 & & \\ a_1 & a_0 & b_1 & b_0 & \\ a_2 & a_1 & b_2 & b_1 & b_0 \\ a_3 & a_2 & & b_2 & b_1 \\ & a_3 & & & b_2 \end{pmatrix}$$
Now recall that $\renewcommand{\s}{\sigma}\det S = \sum_{\s\in S_{d+e}} |\s|\prod s_{i\s(i)}$. I claim that when $s_{ij}\neq 0$, $\deg_x s_{ij} \leq \deg(f)-d+i-j$ or $\deg_x s_{ij} \leq \deg(g)-e+i-j+e=\deg(g)+i-j$ depending on whether $j\leq e$ or not. Therefore $$\deg_x \prod s_{i\s(i)} \leq e(\deg(f)-d)+d(\deg(g)) + \sum i-\s(i).$$ As $\sum i-\s(i) =0$ and $e(\deg(f)-d) + d(\deg(g)) \leq \deg(f)\deg(g)$, we have the result.