Assume we have a curve $X\subseteq \mathbb P^n$ and a hyperplane $H$. Let $x\in X\cap H$. I want to know how to compute the intersection multiplicity $m(X,H,x)$.
As the argument is local, we assume $H = V(x_n)$ and $x$ be the origin in $\mathbb A^n$. Write $X = V(I)$.
I have 2 attempts:
By length definition, $m(X,H,x) = l((k[x_1,\dots,x_n]/(I + (x_n)))_{(x_1,\dots,x_n)})$. But I don't know how to continue computing this length.
Regard $H$ as a Cartier divisor and pullback $H$ to $X$. Then $m(X,H,x) = val_x(x_n)$. But I don't know how to find a local parameter of $(k[x_1,\dots,x_n]/I)_{(x_1,\dots,x_n)}$.
Is there an algorithm to calculate $m(X,H,x)$? At least I want to prove $m(X,H,x) = 1$ if and only if $T_xX\notin H$.
We'll suppose $X$ is a smooth integral curve not contained in $H$. We'll also assume we're working over an algebraically closed field (this is likely what you're assuming, given your remark about assuming the point is the origin).
The statement you're interested in follows from the following two claims.
Claim: $X\cap H$ is nonsingular at $x$ iff $T_xX\not\subset H$.
Proof: locally at $x$, $X\cap H$ is cut out by the local equations for $X$ and the local equation of $H$. Therefore the Jacobian matrix for $X\cap H$ at $x$ is the Jacobian matrix of $X$ at $x$ concatenated with the Jacobian matrix of $H$ at $x$. As the tangent space is the kernel of the Jacobian, we see that if $T_xX\subset H$ iff the kernel of the matrix doesn't change after concatenation. But the kernel is zero-dimensional (i.e. one dimension smaller than the kernel of the Jacobian of $X$) iff $X\cap H$ is nonsingular at $x$.
Claim: $X\cap H$ is nonsingular iff the intersection multiplicity of $X$ and $H$ at $x$ is 1.
Proof: Note that since $X\cap H$ is finite discrete, $X\cap H$ is locally at $x$ the affine scheme $\operatorname{Spec} \mathcal{O}_{X,x}/(h)$, where $h$ is a local equation for $H$. This is regular iff it's $k$, the base field, which occurs iff $h$ has valuation $1$ in $\mathcal{O}_{X,x}$.
Calculating the valuation of given hyperplane at a specified point on a curve can be done, but it requires calculations unique to each curve and each hyperplane (and potentially each way of representing your curve). The above proof is general and addresses what you say you're actually after, so I hope you'll forgive me being brief. If you have specific questions about how to do the calculation in a certain example, it may make more sense to post that as a new question.
More details with a worked example:
Let $X$ be the twisted cubic in $\Bbb P^3$, that is, $V(t_0t_2-t_1^2,t_1t_3-t_2^2,t_0t_3-t_1t_2)$. Let $x=[1:0:0:0]$. Let $H=V(at_1+bt_2+ct_3)$. On the open affine patch $D(t_0)$, $X$ is cut out by $(t_2-t_1^2,t_1t_3-t_2^2,t_3-t_1t_2)$ and $H$ is cut out by $(at_1+bt_2+ct_3)$.
The Jacobian of $X$ on the open patch $D(t_0)$ is $$\begin{pmatrix} -2t_1 & 1 & 0 \\ t_3 & -2t_2 & t_1 \\ -t_2 & -t_1 & 1 \end{pmatrix}$$ and evaluating at $x$ we find the Jacobian is $$\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ which has kernel $\{(t,0,0)\}$, equal to the tangent space of $X$ at $x$.
Now let's look at the Jacobian of $X\cap H$ on the open patch $D(t_0)$ and then evaluate at $x$: $$\begin{pmatrix} -2t_1 & 1 & 0 \\ t_3 & -2t_2 & t_1 \\ -t_2 & -t_1 & 1 \\ a & b & c \end{pmatrix} \mapsto_{t_1=t_2=t_3=0} \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \\ a & b & c \end{pmatrix}$$ The kernel of this matrix is trivial iff $a\neq 0$, and $a=0$ iff the hyperplane $h$ includes the tangent space to our curve $X$ at $x$.
To verify what's happening on the local ring side of things, $\mathcal{O}_{X,x} = k[t_1,t_2,t_3]_{(t_1,t_2,t_3)}/(t_2-t_1^2,t_1t_3-t_2^2,t_3-t_1t_2)$ which is isomorphic to $k[t_1]_{(t_1)}$ via sending $t_2\mapsto t_1^2$ and $t_3\mapsto t_1^3$. Therefore $t_1$ is a uniformizer and $t_i$ has valuation $i$. Our hyperplane $at_1+bt_2+ct_3$ therefore has valuation $1$ iff $a\neq 0$ and valuation $>1$ iff $a=0$. So everything works correctly.