Let me suggest a more geometric approach than the one you are using:
Let $C$ be your degree $d$ plane curve. Choose a point $P$ not on $C$, and not
on the $x$-axis.
Let $\mathbb P^1$ be the $x$-axis.
Now let $\phi:C \to \mathbb P^1$ be projection from $P$ to the $x$-axis, i.e.
given any point $Q$ in $C$, draw a line from $P$ to $Q$, and see where it intersects the $x$-axis; that point is the value $\phi(Q)$.
This will be a map of degree $d$ (if you take a point $R$ in the $x$-axis, and count
how many points lie in its preimage, you draw the line through $P$ and $R$,
and count how many times this intersects $C$; the answer will be $d$ times,
because $C$ is of degree $d$).
Now you can apply the Riemann--Hurwitz formula; the only problem is to work out the ramification points.
Ramification will occur when the line through $P$ and $Q$ has a multiple intersection with $C$ at $Q$, i.e. is tangent to $C$ at $Q$. If you choose everything generically (e.g. choose $P$ generically, and/or change coordinates
so that $C$ is in a general position with regard to the $x$-axis), then these
intersections multiplicities will never be more than $2$ (i.e. you will get tangencies, but never higher order tangencies), and so $e_{\phi}(Q)$ will be either $1$ (if the line through $P$ and $Q$ is not tangent to $C$ at $Q$) or
$2$ (if the line through $P$ and $Q$ is tangent to $C$ at $Q$).
Now you have to figure out how many times a tangency occurs.
Of course, you can work this out by assuming the answer (i.e. that $g(C)
= (d-1)(d-2)/2$, and working backwards). I suggest that you also write
down an explicit conic, like $x^2 + y ^2 = 1$, and then perhaps a higher
degree curve, and concretely apply the above
procedure and see directly how many points of tangency occur. After doing all
this, you will hopefully figure out in general how many $Q$s there are for which the line
through $P$ is tangent to $C$, and also see how to prove that what you
have worked out is correct.
To fill in all the details you will either need to make the "general position"
argument above rigorous, or deal with the possibility of higher order tangency (i.e. the case when $e_{\phi}(Q) > 2$). But I would worry about this later,
after you understand the basic geometry of the situation.
The genus $g$ of a curve $C$ can be defined as follows:
If $C$ is non-singular, then $g=g(C)$ is the number that appears in the Riemann-Roch theorem.
If $C$ is singular, then we find a curve $C'$ that is non-singular, and birational to $C$ (you can find $C'$ by resolution of singularities, for example). In this case, $g(C)$ is defined to be $g(C')$.
In particular, a curve of genus $1$ does not need to be smooth. The fact that $g(C)=1$ only implies that $C$ is birational to a smooth curve $C'$ of genus $1$, and if $C'$ has a rational point, then $C'$ is an elliptic curve and it has a Weierstrass form. Since $C$ and $C'$ are birational, then $C$ is birational to a Weierstrass form (which still does not imply that $C$ is smooth).
I found the answers to this question quite useful here.
Best Answer
The notation $\text{genus}(C_1)$ doesn't even make sense unless you know that the genus is invariant under isomorphism; if it isn't, the genus must depend on information other than $C_1$ which you haven't provided.
In any case, the definition of genus you are given implies that it is unique, and since the various dimensions $\ell(D)$ are defined independently of any choices they are automatically invariant under isomorphism, so the definition you have been given already comes with a guarantee that $g$ is invariant under isomorphism. But if you want a "proof" anyway, then setting $D = 0$ gives $\ell(K_C) = g$, so it suffices to show that the canonical divisor is invariant under isomorphism (that's why it's called the canonical divisor!).