Defining the convolution for finite-length signals
The OP says that sequences

$x = [x(0), …, x(N_x-1)]$ and $h = [h(0), …, h(N_h-1)]$ (are) of different finite lengths $N_x$ and $N_h$ respectively

He does not claim that $x$ and $h$ are finite-length segments of possibly longer (maybe even infinitely longer) sequences, and so the standard assumption is that finite length means what it says. In particular, $x[n]$ has value $0$ whenever $n &lt; 0$ or whenever $n \geq N_x$, and similarly, $h[n]$ has value $0$ whenever $n &lt; 0$ or whenever $n \geq N_h$.
The general definition of the convolution of sequences $p$ and $q$ is that result of the convolution is another sequence, which we denote as $\big(p\star q\big)$ whose $n$-th term is given by
$$\big(p\star q\big)[n] = \sum_{k=-\infty}^\infty p[k]q[n-k] = \sum_{k=-\infty}^\infty p[n-k]q[k]\tag{1}$$
subject to the usual shibboleths about convergence of the sums and the like.  So, applying this to the case at hand, we have that
\begin{align}
\big(x\star h\big)[n] &amp;= \sum_{k=-\infty}^\infty x[k]h[n-k]\\
&amp;= \sum_{k=0}^{N_x-1} x[k]h[n-k] &amp; \scriptstyle{\text{because $x[k]=0$ for $k&lt;0$ or $k\geq N_x$}} \tag{2}
\end{align}
Using the alternative formula for $\big(p\star q\big)[n]$, we have
\begin{align}
\big(x\star h\big)[n] &amp;= \sum_{k=-\infty}^\infty x[n-k]h[k]\\
&amp;= \sum_{k=0}^{N_h-1} x[n-k]h[k] &amp; \scriptstyle{\text{because $h[k]=0$ for $k&lt;0$ or $k\geq N_x$}} \tag{3}
\end{align}
Both $(2)$ and $(3)$ are valid expressions for $\big(x\star h\big)[n]$ and for each integer $n$, both give the same value for $\big(x\star h\big)[n]$ even though the expressions are seemingly quite different.. The OP claims that
$$\sum_{i = 0}^{N_x-1} x(i) h(n - i) \neq \sum_{i = 0}^{N_h-1} h(i) x(n - i)$$
but this claim is false; for every choice of integer $n$, the two sums result in the same expression (except for a trivial re-arrangement of the terms).  Don't believe it?  Let's try a few special cases keeping firmly in mind that $x[n]$ and $h[n]=0$ if $n &lt; 0$.
For $n = 0$,
\begin{align}\require{cancel}
\sum_{k=0}^{N_x-1} x[k]h[0-k] &amp;= x[0]h[0] + x[1]\cancelto0{h[-1]} + x[2]\cancelto0{h[-2]} + \cdots\\
&amp;= x[0]h[0]\\
\sum_{k=0}^{N_h-1} x[0-k]h[k] &amp;= x[0]h[0] + \cancelto0{x[-1]}h[1] + \cancelto0{x[-2]}h[2] + \cdots \\&amp;= x[0]h[0]
\end{align}
For $n = 1$,
\begin{align}
\sum_{k=0}^{N_x-1} x[k]h[1-k] &amp;= x[0]h[1] + x[1]h[0] + x[2]\cancelto0{h[-1]} + \cdots \\&amp;= x[0]h[1]+x[1]h[0]\\
\sum_{k=0}^{N_h-1} x[1-k]h[k] &amp;= x[1]h[0] + x[0]h[1] + \cancelto0{x[-1]}h[2] + \cdots \\&amp;= x[1]h[0] + x[0]h[1]
\end{align}
More generally, assuming that $N_h &lt; N_x$, for each $n$, $0 \leq n \leq N_h -1$,
$$\sum_{k=0}^{N_x-1} x[k]h[n-k] = x[0]h[n] + x[1]h[n-1] + \cdots + x[n]h[0]\tag{4}$$
while
$$\sum_{k=0}^{N_h-1} x[n-k]h[k] = x[n]h[n]0] + x[n-1]h[1] + \cdots + x[0]h[n]\tag{5}$$
which is the same sum as in $(4)$ except for being in reverse order! Fortunately, addition is commutative.....
That the sums in $(2)$ and $(3)$ are equal (except for being written in reverse order) is true not just for $0 \leq n \leq N_h -1$ but for all $n$. I won't bother typing out the gory details, but each sum in $(2)$ and $(3)$ contains
at most $\min(N_x,N_h)$ nonzero terms and they occur in reverse order one from the other.  $(2)$ and $(3)$ are the same sum even though the expressions look different at first glance, regardless of the OP's claims, and convolution is indeed commutative: for every integer $n$,
$$\big(p\star q\big)[n] = \big(q\star p\big)[n].$$