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The central idea is that you can translate probabilities from a single-particle picture into fractions of particles in a many-particle picture (assuming no interaction).

Consider $T$ for a moment for a single particle. Let's just, for ease of writing, say $T=.9$. So if you send in a single particle from the far left, 90% of the probability density continues moving right, and 10% of the probability density bounces back. So if you have a perfect detector at the far right, there is a 90% chance your detector will go off.

Now consider doing the same thing with a whole beam of particles (that don't interact with each other or in any way effect each other's propagation through the disturbance). If you send in $N$ particles, you will expect the detector on the right to go off $.9N$ times, because each particle individually had a 90% chance of hitting it. So, on the far right, you could say you could meaningfully say you detect 90% of the particles you sent in.

In that sense:

$$T=\frac{|j_\mathrm{trans}|}{|j_\mathrm{inc}|}=\frac{N|j_\mathrm{trans}|}{N|j_\mathrm{inc}|}=\frac{\text{num transmitted}}{\text{num incident}}$$

Edit to explain: "Why is $T$ the probability of a particle being transmitted?"

$T$ and $R$ are normally defined as a function of energy $E$ of the incoming wave. How do replace statements about incoming waves with statements about particles?

In quantum mechanics, we think of "an incoming particle" as a right-moving wavepacket from the left. (A wavepacket of energy ~$E$ is a superposition of waves of a small range of frequencies tight around $E/\hbar$, so the resulting wavefunction is not spread over all of position space, but rather is normalizable and concentrated in some region, and still has energy of roughly $E$.) So it "looks like a particle going right."

If we have a tight wavepacket so that $T(E)$ is a constant $T$ over the frequencies of interest, then as you evolve the system, 90% of that wavepacket will end up going right past the disturbance, and 10% of the wavepacket will get reflected back. It's easier to visualize this with an animation.

Griffiths discusses this topic at the end of Sec 2.5 (pg 75-76 in my copy). Shankar mentions it in Sec 5.4 (pg 175 for me), but the math of showing this apparently gets quite involved, so neither author puts forward a real proof I'm afraid. But see this related question. (Note that the notation in that question is different: they use $T$ and $R$ for the coefficients rather than the probabilities as you've done. So your $T,R$ is their $|T|^2/|A|^2,|R|^2/|A|^2$.)

Edit 2: To explain why we have ratios

The ratios are only there because when we work the problem with incoming and outgoing plane waves, our state is not normalized. If you (hypothetically) work the problem with a normalized wavepacket, then the total integrated probability density coming in from the far left (before reflection) is 1. Then the total integrated probability density moving off to the right at the end is just $T=\int_{t_i}^{t_f} j_\mathrm{trans}dt$ and this is, by definition, the probability of the particle transmitting.

When actually work the problem in practice, however, you generally don't use wavepackets, you use incoming and outgoing waves, which are not normalized. The total integrated probability density flowing in is thus not 1, but rather $j_\mathrm{inc}\times \mathrm{time}$, and the total integrated probability density flowing out is $j_\mathrm{trans}\times \mathrm{time}$.

The question we want to answer is "What would be the probability density flowing out if the total integrated incoming density is 1?" So we divide our naive unnormalized $T=j_\mathrm{trans}$ by $I=j_\mathrm{inc}$ to fix the fact that we didn't normalize our state from the beginning.

The central idea is that you can translate probabilities from a single-particle picture into fractions of particles in a many-particle picture (assuming no interaction).

Consider $T$ for a moment for a single particle. Let's just, for ease of writing, say $T=.9$. So if you send in a single particle from the far left, 90% of the probability density continues moving right, and 10% of the probability density bounces back. So if you have a perfect detector at the far right, there is a 90% chance your detector will go off.

Now consider doing the same thing with a whole beam of particles (that don't interact with each other or in any way effect each other's propagation through the disturbance). If you send in $N$ particles, you will expect the detector on the right to go off $.9N$ times, because each particle individually had a 90% chance of hitting it. So, on the far right, you could say you could meaningfully say you detect 90% of the particles you sent in.

In that sense:

$$T=\frac{|j_\mathrm{trans}|}{|j_\mathrm{inc}|}=\frac{N|j_\mathrm{trans}|}{N|j_\mathrm{inc}|}=\frac{\text{num transmitted}}{\text{num incident}}$$

Edit to explain: "Why is $T$ the probability of a particle being transmitted?"

$T$ and $R$ are normally defined as a function of energy $E$ of the incoming wave. How do replace statements about incoming waves with statements about particles?

In quantum mechanics, we think of "an incoming particle" as a right-moving wavepacket from the left. (A wavepacket of energy ~$E$ is a superposition of waves of a small range of frequencies tight around $E/\hbar$, so the resulting wavefunction is not spread over all of position space, but rather is normalizable and concentrated in some region, and still has energy of roughly $E$.) So it "looks like a particle going right."

If we have a tight wavepacket so that $T(E)$ is a constant $T$ over the frequencies of interest, then as you evolve the system, 90% of that wavepacket will end up going right past the disturbance, and 10% of the wavepacket will get reflected back. It's easier to visualize this with an animation.

Griffiths discusses this topic at the end of Sec 2.5 (pg 75-76 in my copy). Shankar mentions it in Sec 5.4 (pg 175 for me), but the math of showing this apparently gets quite involved, so neither author puts forward a real proof I'm afraid. But see this related question. (Note that the notation in that question is different: they use $T$ and $R$ for the coefficients rather than the probabilities as you've done. So your $T,R$ is their $|T|^2/|A|^2,|R|^2/|A|^2$.)

Edit 2: To explain why we have ratios

The ratios are only there because when we work the problem with incoming and outgoing plane waves, our state is not normalized. If you (hypothetically) work the problem with a normalized wavepacket, then the total integrated probability density coming in from the far left (before reflection) is 1. Then the total integrated probability density moving off to the right at the end is just $T=\int_{t_i}^{t_f} j_\mathrm{trans}dt$ and this is, by definition, the probability of the particle transmitting.

When actually work the problem in practice, however, you generally don't use wavepackets, you use incoming and outgoing waves, which are not normalized. The total integrated probability density flowing in is thus not 1, but rather $j_\mathrm{inc}\times \mathrm{time}$, and the total integrated probability density flowing out is $j_\mathrm{trans}\times \mathrm{time}$.

The question we want to answer is "What would be the probability density flowing out if the total integrated incoming density is 1?" So we divide our naive unnormalized $T=j_\mathrm{trans}$ by $I=j_\mathrm{inc}$ to fix the fact that we didn't normalize our state from the beginning.

The central idea is that you can translate probabilities from a single-particle picture into fractions of particles in a many-particle picture (assuming no interaction).

Consider $T$ for a moment for a single particle. Let's just, for ease of writing, say $T=.9$. So if you send in a single particle from the far left, 90% of the probability density continues moving right, and 10% of the probability density bounces back. So if you have a perfect detector at the far right, there is a 90% chance your detector will go off.

Now consider doing the same thing with a whole beam of particles (that don't interact with each other or in any way effect each other's propagation through the disturbance). If you send in $N$ particles, you will expect the detector on the right to go off $.9N$ times, because each particle individually had a 90% chance of hitting it. So, on the far right, you could say you could meaningfully say you detect 90% of the particles you sent in.

In that sense:

$$T=\frac{|j_\mathrm{trans}|}{|j_\mathrm{inc}|}=\frac{N|j_\mathrm{trans}|}{N|j_\mathrm{inc}|}=\frac{\text{num transmitted}}{\text{num incident}}$$

Edit to explain: "Why is $T$ the probability of a particle being transmitted?"

$T$ and $R$ are normally defined as a function of energy $E$ of the incoming wave. How do replace statements about incoming waves with statements about particles?

In quantum mechanics, we think of "an incoming particle" as a right-moving wavepacket from the left. (A wavepacket of energy ~$E$ is a superposition of waves of a small range of frequencies tight around $E/\hbar$, so the resulting wavefunction is not spread over all of position space, but rather is normalizable and concentrated in some region, and still has energy of roughly $E$.) So it "looks like a particle going right."

If we have a tight wavepacket so that $T(E)$ is a constant $T$ over the frequencies of interest, then as you evolve the system, 90% of that wavepacket will end up going right past the disturbance, and 10% of the wavepacket will get reflected back. It's easier to visualize this with an animation.

Griffiths discusses this topic at the end of Sec 2.5 (pg 75-76 in my copy). Shankar mentions it in Sec 5.4 (pg 175 for me), but the math of showing this apparently gets quite involved, so neither author puts forward a real proof I'm afraid. But see this related question. (Note that the notation in that question is different: they use $T$ and $R$ for the coefficients rather than the probabilities as you've done. So your $T,R$ is their $|T|^2/|A|^2,|R|^2/|A|^2$.)

The central idea is that you can translate probabilities from a single-particle picture into fractions of particles in a many-particle picture (assuming no interaction).

Consider $T$ for a moment for a single particle. Let's just, for ease of writing, say $T=.9$. So if you send in a single particle from the far left, 90% of the probability density continues moving right, and 10% of the probability density bounces back. So if you have a perfect detector at the far right, there is a 90% chance your detector will go off.

Now consider doing the same thing with a whole beam of particles (that don't interact with each other or in any way effect each other's propagation through the disturbance). If you send in $N$ particles, you will expect the detector on the right to go off $.9N$ times, because each particle individually had a 90% chance of hitting it. So, on the far right, you could say you could meaningfully say you detect 90% of the particles you sent in.

In that sense:

$$T=\frac{|j_\mathrm{trans}|}{|j_\mathrm{inc}|}=\frac{N|j_\mathrm{trans}|}{N|j_\mathrm{inc}|}=\frac{\text{num transmitted}}{\text{num incident}}$$

Edit to explain: "Why is $T$ the probability of a particle being transmitted?"

$T$ and $R$ are normally defined as a function of energy $E$ of the incoming wave. How do replace statements about incoming waves with statements about particles?

In quantum mechanics, we think of "an incoming particle" as a right-moving wavepacket from the left. (A wavepacket of energy ~$E$ is a superposition of waves of a small range of frequencies tight around $E/\hbar$, so the resulting wavefunction is not spread over all of position space, but rather is normalizable and concentrated in some region, and still has energy of roughly $E$.) So it "looks like a particle going right."

If we have a tight wavepacket so that $T(E)$ is a constant $T$ over the frequencies of interest, then as you evolve the system, 90% of that wavepacket will end up going right past the disturbance, and 10% of the wavepacket will get reflected back. It's easier to visualize this with an animation.

Griffiths discusses this topic at the end of Sec 2.5 (pg 75-76 in my copy). Shankar mentions it in Sec 5.4 (pg 175 for me), but the math of showing this apparently gets quite involved, so neither author puts forward a real proof I'm afraid. But see this related question. (Note that the notation in that question is different: they use $T$ and $R$ for the coefficients rather than the probabilities as you've done. So your $T,R$ is their $|T|^2/|A|^2,|R|^2/|A|^2$.)

Edit 2: To explain why we have ratios

The ratios are only there because when we work the problem with incoming and outgoing plane waves, our state is not normalized. If you (hypothetically) work the problem with a normalized wavepacket, then the total integrated probability density coming in from the far left (before reflection) is 1. Then the total integrated probability density moving off to the right at the end is just $T=\int_{t_i}^{t_f} j_\mathrm{trans}dt$ and this is, by definition, the probability of the particle transmitting.

When actually work the problem in practice, however, you generally don't use wavepackets, you use incoming and outgoing waves, which are not normalized. The total integrated probability density flowing in is thus not 1, but rather $j_\mathrm{inc}\times \mathrm{time}$, and the total integrated probability density flowing out is $j_\mathrm{trans}\times \mathrm{time}$.

The question we want to answer is "What would be the probability density flowing out if the total integrated incoming density is 1?" So we divide our naive unnormalized $T=j_\mathrm{trans}$ by $I=j_\mathrm{inc}$ to fix the fact that we didn't normalize our state from the beginning.

The central idea is that you can translate probabilities from a single-particle picture into fractions of particles in a many-particle picture (assuming no interaction).

Consider $T$ for a moment for a single particle. Let's just, for ease of writing, say $T=.9$. So if you send in a single particle from the far left, 90% of the probability density continues moving right, and 10% of the probability density bounces back. So if you have a perfect detector at the far right, there is a 90% chance your detector will go off.

Now consider doing the same thing with a whole beam of particles (that don't interact with each other or in any way effect each other's propagation through the disturbance). If you send in $N$ particles, you will expect the detector on the right to go off $.9N$ times, because each particle individually had a 90% chance of hitting it. So, on the far right, you could say you could meaningfully say you detect 90% of the particles you sent in.

In that sense:

$$T=\frac{|j_\mathrm{trans}|}{|j_\mathrm{inc}|}=\frac{N|j_\mathrm{trans}|}{N|j_\mathrm{inc}|}=\frac{\text{num transmitted}}{\text{num incident}}$$

Edit to explain: "Why is $T$ the probability of a particle being transmitted?"

$T$ and $R$ are normally defined as a function of energy $E$ of the incoming wave. How do replace statements about incoming waves with statements about particles?

In quantum mechanics, we think of "an incoming particle" as a right-moving wavepacket from the left. (A wavepacket of energy ~$E$ is a superposition of waves of a small range of frequencies tight around $E/\hbar$, so the resulting wavefunction is not spread over all of position space, but rather is normalizable and concentrated in some region, and still has energy of roughly $E$.) So it "looks like a particle going right."

If we have a tight wavepacket so that $T(E)$ is a constant $T$ over the frequencies of interest, then as you evolve the system, 90% of that wavepacket will end up going right past the disturbance, and 10% of the wavepacket will get reflected back. It's easier to visualize this with an animation. Griffiths discusses this topic at the end of Sec 2.5 (pg 75-76 in my copy). Shankar mentions it in Sec 5.4 (pg 175 for me), but the math of showing this apparently gets quite involved, so neither author puts forward a real proof I'm afraid.

The central idea is that you can translate probabilities from a single-particle picture into fractions of particles in a many-particle picture (assuming no interaction).

Consider $T$ for a moment for a single particle. Let's just, for ease of writing, say $T=.9$. So if you send in a single particle from the far left, 90% of the probability density continues moving right, and 10% of the probability density bounces back. So if you have a perfect detector at the far right, there is a 90% chance your detector will go off.

Now consider doing the same thing with a whole beam of particles (that don't interact with each other or in any way effect each other's propagation through the disturbance). If you send in $N$ particles, you will expect the detector on the right to go off $.9N$ times, because each particle individually had a 90% chance of hitting it. So, on the far right, you could say you could meaningfully say you detect 90% of the particles you sent in.

In that sense:

$$T=\frac{|j_\mathrm{trans}|}{|j_\mathrm{inc}|}=\frac{N|j_\mathrm{trans}|}{N|j_\mathrm{inc}|}=\frac{\text{num transmitted}}{\text{num incident}}$$

Edit to explain: "Why is $T$ the probability of a particle being transmitted?"

$T$ and $R$ are normally defined as a function of energy $E$ of the incoming wave. How do replace statements about incoming waves with statements about particles?

In quantum mechanics, we think of "an incoming particle" as a right-moving wavepacket from the left. (A wavepacket of energy ~$E$ is a superposition of waves of a small range of frequencies tight around $E/\hbar$, so the resulting wavefunction is not spread over all of position space, but rather is normalizable and concentrated in some region, and still has energy of roughly $E$.) So it "looks like a particle going right."

If we have a tight wavepacket so that $T(E)$ is a constant $T$ over the frequencies of interest, then as you evolve the system, 90% of that wavepacket will end up going right past the disturbance, and 10% of the wavepacket will get reflected back. It's easier to visualize this with an animation.

Griffiths discusses this topic at the end of Sec 2.5 (pg 75-76 in my copy). Shankar mentions it in Sec 5.4 (pg 175 for me), but the math of showing this apparently gets quite involved, so neither author puts forward a real proof I'm afraid. But see this related question. (Note that the notation in that question is different: they use $T$ and $R$ for the coefficients rather than the probabilities as you've done. So your $T,R$ is their $|T|^2/|A|^2,|R|^2/|A|^2$.)