Question about why this hyperplane section divisor is singular at a certain point I have a small question about an excerpt from a passage about intersection numbers in "Basic Algebraic Geometry I" by Igor Shafarevich.

Let $ X \subset \mathbb{P}^{3} $ be a nonsingular surface of degree $ m $ and $ L \subset X $ a line. Consider a plane in $ \mathbb{P}^{3} $ containing $ L $ and not tangent to $ X $ at at least one point of $ L, $ and let $ E $ be the hyperplane section of $ X $ by this plane. Then $ L $ is contained in $ E $ as a component of multiplicity $ 1 $ and $$ E = L + C \;\text{ with } C = \sum k_{i}C_{i} \; \text{ and } \sum k_{i}\text{deg}C_{i} = m-1.  $$
Observe that the curve $ E $ is singular at a point of intersection of $ L $ and $ C, $ which means that the plane cutting out $ E $ equals the tangent plane to $ X $ at this point. 

I don't quite see why $ E $ is singular at a point $ y \in L \cap C. $  Also, why does this make $ E $ the same as the tangent plane to $ X $ at $ y? $ Does this follow from the singularity of $ y $ by definition? I feel I'm forgetting some basic property.

$E$ is singular at $y\in L\cap C$ because $y$ is on multiple irreducible components: every point which is on multiple irreducible components is singular, as the local ring of such a point has at least two minimal primes (corresponding to the distinct irreducible components it's on) while a regular local ring is a domain and thus has only one minimal prime.
As for the business about the tangent plane, recall that as $X$ is nonsingular, it has a tangent plane at each point. The inclusion of any subvariety $i:Y\hookrightarrow X$ induces an inclusion of tangent spaces $Di_p:T_pY\hookrightarrow T_pX$ at any point $p\in Y$. Let $P$ be the plane cutting out $E$. Then we have the following two inclusions of tangent spaces: $T_yE\hookrightarrow T_yP$ and $T_yE\hookrightarrow T_yX$. As $T_yE$ is at least two-dimensional because the 1-dimensional subscheme $E$ is singular at $y$ and $T_yX$ is two-dimensional as $X$ is smooth at $y$, we see that inclusion must be an isomorphism. Similarly, as $T_yE$ and $T_yP$ are both two-dimensional, that map must be an isomorphism as well. So $T_yE$ is simultaneously the tangent plane to $X$ at $y$ and the plane which cuts out $E$.
