I think you can avoid using the word polar but still get an easy solution. Let's suppose we have a smooth curve in $\mathbb{P}^2$, then we want to count the tangent line to our curve from $P=[r,s,t]$. We can use homogenous coordinates $[x,y,z]$ and our curve is defined by a degree d homogenous polynomial $F(x,y,z)=0$. Imposing that the tangent line through a point of our curve passes through P we get $r\frac{\partial{F}}{\partial{x}}+s\frac{\partial{F}}{\partial{y}}+t\frac{\partial{F}}{\partial{z}}=0$, which is a homogenous polynomial of degree $d-1$, hence a curve. By Bezout the new curve and our original curve intersect in a finite numbers of points ($d(d-1)$) and from our construction we see that the points of intersection of the two curves are exactly the points whose tangent passes through P, hence these points are finite.
If you don't like this approach you can use Bertini and in this case I wouldn't say it is an overkill.
We have a curve $C$, which is the zero set of some equation $F$ (assume $F(0,0)=0$, so $C$ passes through the origin).
If we have a vector $v\in k^n$, we can talk about the line through the origin in the direction $v$, $\ell = \{tv : t\in k\}$, and we can discuss what $F$ restricted to that line looks like.
Well, if we are working in the plane, $v = (x,y)$ so the line is $(tx,ty)$, and $$F|_L(t)=F_m(tx,ty) + F_{m+1}(tx,ty)+\cdots +F_n(tx,ty).$$
Since each of the forms $F_i$ is homogeneous, we can pull the $t$s out to get
$$F|_L(t) = F_m(x,y)t^m+F_{m+1}(x,y)t^{m+1}+\cdots +F_n(x,y)t^n.$$
Then if we ask about the multiplicity of the zero of $F|_L$ at $0$, we find that it is the first $i$ such that $F_i(x,y)\ne 0$. For most choices of $x$ and $y$, we will have $F_m(x,y)\ne 0$, so for most lines, the multiplicity of the zero of $F|_L$ at the origin is $m$, the same as the multiplicity of $F$ at the origin. However for lines in directions $(x,y)$ where $F_m(x,y)=0$, the multiplicity goes up. These directions correspond to tangent directions, since when you travel in a tangent direction, the function defining an implicit surface should change more slowly or have an extra zero (since the tangent directions should intuitively correspond to infinitesimal movements that change the value of the function the least).
Then, in particular, in two dimensions, we can use the fact that every homogeneous polynomial in two variables over an algebraically closed field splits into linear factors. Each of these factors will correspond to one of the direction vectors $(x,y)$ for which $F_m(x,y)=0$. Thus, we get that the tangent directions in two dimensions are determined by the linear factors of the lowest degree form.
To focus on your specific question, the reason the low degree form is what matters is that the lowest degree form has to be zero for the multiplicity of a zero to be greater than $m$.
Best Answer
Let $k=\mathbb{C}.$ I think, the best intuitive explanation is related to the notion of "tangent cone" (See Gathmann's notes page 61) which is the "best" approximation of $C$ around a singularity.
You know that the linear part of $F$ is the simplest approximation of $C$ around $P,$ so in the absence of the linear part (singularity), we should stick to the second option which is the non-zero homogeneous part $F_m$ of $F$ with the smallest degree in the homogeneous decomposition of $F.$ (for example, let $F(X,Y)=(Y^2-X^2)-X^3$ then $F_0=F_1=0,$ $F_2=Y^2-X^2$ and $F_3=-X^3.$)
The affine algebraic set generated by the ideal $(F_m)$ forms a cone over the singularity $P$ and is called the tangent cone. (in the above example, it is $V(Y^2-X^2)\subset \mathbb{A}^2$ which contains two tangent lines $Y=\pm X$ of $C$ at $P.$) Then, if $F_m$ factorizes into linear products $\lambda\prod_{i=1}^m(X-\alpha_iY)$ the affine zero locus is the tangent cone at $P$ contianing $m$ tangent lines of $C$ at $P.$ Of course, since $\alpha_i$'s may not be distinct in general, you may have a sort of thickened tanget lines (for example, $F(X,Y)=Y^3-X^2$ then $F_2=-X^2$ and $V(X^2)$ is the thickened $x$-axis.)
Furthermore, the projectivized tangent cone i.e. $V_p(F_m)$ where $V_p(I)$ is the projective zero locus of $I$ is the exceptional locus in the blowup of $C$ at $P.$ (in our example, then the exceptional locus of $\tilde{C}$ the blowup of $C$ at $P,$ has two points corresponding to two tangent lines $Y=\pm X.$ )