Why, in simple terms, is Cohen's D not affected by sample size but a T-Test is? Why, in simple terms, is Cohen's D not affected by sample size but a T-Test is? I don't really understand and tbh am struggling to tell the difference between the two at all.

Standardized mean difference is the raw effect size (ie difference of the means of two populations) divided by the standard deviation of the population (in the simple case when both population have the same deviation). It is a way to express the effect size in a dimensionless way.
The standardized mean difference, is a property of the populations. It has not to do with the sample size and is not influenced by the sample size.
Cohen's $\bf d$ is a statistic that estimates the standardized mean difference.
For Cohen's $d$, which is a statistic computed with a sample, there will be an influence of the sample size on the distribution of the statistic. This difference is that for increasing sample size the Cohen's $d$ will be more close to the standardized mean difference (but the expectation value of Cohen's $d$ will not be affected by the sample size).
The t-statistic is the observed effect size divided by the standard error of the observed effect size. This error becomes smaller for larger samples, and the t-statistic will be larger for smaller samples (provided the effect size is non-zero).

Example
When we estimate the difference in the means of two populations, $\mu_2- \mu _1$, then we have the following expressions for the standard mean difference, Cohen's $d$, and the t-statistic.
Standardized mean difference is a property of the populations and independent from $n$ and just the difference in the (true) population means divided by the (true) standard deviation of the populations (more complex formula's exist when the population's standard deviations are unequal)
$$SMD = \frac{\mu_2- \mu_1}{\sigma}$$
Below is an example for two Gaussian distributed populations. The effect size is in this case two standard deviations so $SMD=2$.

Cohen's $\bf d$ is a statistic that estimates the standardized mean difference from the sample in which case you have
$${d} = \frac{\bar{x}_1-\bar{x}_2}{s}$$
with the sample means $\bar{x}_i$, and $s$ the estimate of the distribution standard deviations (for the simple case that the deviations are assumed to be equal, $s$ is the pooled standard deviation).
The t-statistic
$$t =  \frac{\bar{x}_2-\bar{x}_1}{S.E.(\bar{x}_2-\bar{x}_1)} = \frac{\bar{x}_2-\bar{x}_1}{s \cdot \sqrt{2/n}} = \sqrt{n/2} d
$$
This standard error of the difference between the sample means, $S.E.(\bar{x}_2-\bar{x}_1) = s \cdot \sqrt{2/n}$, scales with the sample size and becomes smaller for larger samples.

See below an example of the distribution of ${d}$ and $t$ for different sample sizes when the true standardized effect size is $SMD=0.1$. (the distributions will be non-central t-distributions)

For a given effect size, when ${n}$ increases then the t-statistic can become very high. This is because the error of the observed effect size becomes smaller.
On the other hand, Cohen's $d$, will approach the real effect size (standardized mean difference) and does not become bigger for increasing sample size $n$.
Even very small effect sizes (in terms of the standardized mean difference) can result in a large t-statistic as long as $\sqrt{n}$ (the sample size) is large enough.
