This is only weird because you have slipped into the unphysical limit where the mass of the string is zero.
Let's abstract away the pulley, and imagine a horizontal system where there's a massive rod connecting the two masses:
,-, ,---,
| |---------------------| |
`-' `---'
m µ M
We can arrange for $m$ to be pulled to the left by a force $mg$ and for $M=2m$ to be pulled to the right by $Mg$. There is nothing magic about the weights of the objects; we can just pull on them that hard horizontally with our arms or our students or something.
Now our equations of motion are
\begin{align}
-mg + T_\text{left} &= ma \tag1
\\
-T_\text{left} + T_\text{right} &= \mu a \tag2
\\
-T_\text{right} + Mg &= Ma \tag3
\end{align}
If we add these three equations, we see that the internal forces cancel and that the whole system is accelerating as a unit in response to the external forces:
\begin{align}
-mg + (T_\text{left} - T_\text{left}) + (T_\text{right}-T_\text{right}) + Mg &= (m + \mu + M)a
\\
-mg + Mg &= (m + \mu + M)a \tag4
\end{align}
In a typical lab experiment, we might have $m\approx\rm1\,kg$, $M\approx\rm2\,kg$, and $\mu\approx\rm1\,g$. The fourth equation tells us that, with these mass ratios, we'll get pretty much the correct acceleration for the combined system if we approximate $\frac{\mu}{m+M} \approx 10^{-3}$ as zero: in fact, neglecting $\mu$ introduces a smaller rounding error than disagreeing about whether to use $g \approx \rm 9.80\,m/s^2$ versus $g \approx \rm 9.81\,m/s^2$.
Likewise your result $T_\text{left} = T_\text{right} = \frac43 mg$ is still going to be good to three or four significant figures, even though we snuck in the low-mass string.
In fact, our second equation gives the difference between the two tension forces as
$$
-T_\text{left} + T_\text{right}
=
\mu g \frac{M-m}{m + \mu + M}
≈ \frac 13 \mu g
$$
so the exact result is going to be something like
$$
T_\text{right} = \frac43 mg + \text{(one-ish)}\times \mu g
$$
where we could find the correction factor if we wanted to, but it's clearly still a small correction.
The string doesn't "transmit" the tension. The string is subject to forces on each end, and the nonzero difference between these forces is required to accelerate the string in tandem with the masses. But because the string's mass is so small, only a tiny fraction of the net external force is "expended" in accelerating the string, and pretending the string isn't there at all is a useful simplifying assumption.
So, the block $m$ applied a force greater than it could on the string? How is that possible?
Nope. The left block and string apply $T_\text{left}$ on each other, and the difference between $mg$ and $T_\text{left}$ is what makes that block accelerate.
