Let's examine how points in the exceptional divisor correspond to tangent directions. If $P_1=[0:0:1]$ and we switch to affine coordinates for simplicity, then a curve $C$ through $P_1$ which has the tangent vector $(a,b)\in T_{P_1}\Bbb P^2$ can be written as $V(bx-ay+\cdots)$ (with, WLOG, $a\neq 0$) for all terms of $\cdots$ of degree at least 2 (see for instance here for an algebraic geometry explanation, or maybe think back to calculus for some motivation in terms of derivatives and things like that).
Writing $t,u$ as coordinates on $\Bbb P^1$ so that $Bl_{P_1}\Bbb P^2 = \{([x:y:z],[t:u]) \mid xu=ty\}\subset\Bbb P^2\times\Bbb P^1$, the strict transform of our curve will meet the exceptional divisor inside the chart where $z=1$ and $t=1$. Writing this chart as $\operatorname{Spec} k[x,y,u]/(xu-y)\cong \operatorname{Spec} k[x,u]$ mapping down to the standard affine neighborhood of $P_1$ by $x=x$, $y=xu$, we have that our curve has preimage $V(bx-axu+\cdots)$ with $\cdots$ having terms of degree at least two in $x$. Dividing out by $x$ to get the strict transform, we find that's given by $V(b-au+\cdots)$ where everything in $\cdots$ has positive degree in $x$. Restricting to the exceptional divisor given by $V(x)$, we get $V(b-au)\subset\Bbb P^1$ which is exactly $[a:b]$.
On the other hand, tangent directions to $P_1$ also correspond to length-two subschemes supported at $P_1$ (see for instance some previous discussion on MSE here). We can make the connection explicit: given a tangent vector $(a,b)\in T_{P_1}\Bbb P^2$, this corresponds to the length-two subscheme $\operatorname{Spec} k[x,y]/(x^2,xy,y^2,bx-ay)$, and indeed intersecting our curve $C$ with $V(\mathfrak{m}_{P_1}^2)\subset \Bbb A^2$ gives exactly that: $V(bx-ay+\cdots) \cap V((x,y)^2) = V(bx-ay,x^2,xy,y^2)$.
Connecting an infinitely near point in the exceptional divisor to a tangent direction and then to the length-two subscheme, we get a way to think about what it means for $(L_{12}+L_{34})\cap (L_{13}+L_{14}) = \{P_1,P_2,P_3,P_4\}$. For instance, if we choose coordinates so that $P_1=[0:0:1]$, $P_2$ is the tangent direction $(1,0)$, and $P_3,P_4$ are in $D(z)$, in affine coordinates we get that $L_{12}=V(y)$, $L_{34}=V(\ell_{34})$ where $\ell_{34}$ has nonzero constant term since $P_1,P_3,P_4$ are not collinear, and then $L_{13}=V(\ell_{13})$ and $L_{14}=V(\ell_{14})$ where $\ell_{13}$ and $\ell_{14}$ have no constant term since they vanish at $P_1$ and the coefficient on $x$ in $\ell_{13}$ and $\ell_{14}$ is nonzero since $P_3,P_4$ are not on the line $P_1P_2$. Computing the intersection, we get a length-two subscheme supported at $P_1$, and $P_3$ and $P_4$ appearing as reduced points.
Which length-two subscheme is it? It has coordinate algebra $k[x,y]_{(x,y)}/(y\cdot\ell_{34},\ell_{13}\ell_{14})$; since $\ell_{34}$ has nonzero constant term, it's a unit in $k[x,y]_{(x,y)}$, and $(y\cdot\ell_{34},\ell_{13}\ell_{14})=(y,\ell_{13}\ell_{14})$, which is equal to $(y,x^2)$. Rewriting, $k[x,y]_{(x,y)}/(y,x^2)\cong k[x,y]/(y,x^2) = k[x,y]/(y,x^2,xy,y^2)$, so it's the length-two subscheme corresponding to the tangent direction $(1,0)$. And it's reasonable to call this $\{P_1,P_2\}$, which is an interpretation of what Hartshorne is doing here.
PS: this is not really very well explained in Hartshorne, and I don't know where it is explained well - I picked this up in graduate school based off the introduction to a seminar talk by a visiting speaker. One other frustrating thing is that as you noticed, not all the interpretations of infinitely near points play nice with taking an intersection - the intersections of the strict transforms of $L_{12}+L_{34}$ and $L_{13}+L_{14}$ in $Bl_{P_1}\Bbb P^1$ only gives you two points, since the strict transforms of $L_{12}$, $L_{13}$, and $L_{14}$ all intersect the exceptional divisor in distinct points by the no-3-points-collinear assumption.
