#JavaScript (ES6), 109 bytes

This is really just a port of Delfad0r's great answer, implemented as nested recursive functions.

Takes input as (n)(m).

n=>m=>(g=i=>i?[...g(i-1),(h=k=>k&&h(k-1)-(b=(y,x=k)=>y?x*b(y-1,x-1)/y:i-k&1||-1)(k,i)*b(n)*b(m))(i)]:[])(n+m) 

Try it online!

###How?

The function \$b(y,x)\$ computes:

$$-(-1)^{i-k}{x\choose y}$$

It's slightly shorter to return \$-(-1)^{i-k}\$ as the last iteration of this function than processing it separately. Because we compute the product of three binomial coefficients, the signs of the first two of them are simply cancelling each other out.

The function \$h()\$ is used to compute:

$$\sum_{k=1}^{i}-b(k,i)\times b(n,k)\times b(m,k)\\ =\sum_{k=1}^{i}{(-1)^{i-k}{i\choose k}{k\choose n}{k\choose m}}$$

And the function \$g()\$ is used to compute each term \$a_i\$ for \$1\le i \le n+m\$.

JavaScript (ES6), 109 bytes

This is really just a port of Delfad0r's great answer, implemented as nested recursive functions.

Takes input as (n)(m).

n=>m=>(g=i=>i?[...g(i-1),(h=k=>k&&h(k-1)-(b=(y,x=k)=>y?x*b(y-1,x-1)/y:i-k&1||-1)(k,i)*b(n)*b(m))(i)]:[])(n+m) 

Try it online!

How?

The function \$b(y,x)\$ computes:

$$-(-1)^{i-k}{x\choose y}$$

It's slightly shorter to return \$-(-1)^{i-k}\$ as the last iteration of this function than processing it separately. Because we compute the product of three binomial coefficients, the signs of the first two of them are simply cancelling each other out.

The function \$h()\$ is used to compute:

$$\sum_{k=1}^{i}-b(k,i)\times b(n,k)\times b(m,k)\\ =\sum_{k=1}^{i}{(-1)^{i-k}{i\choose k}{k\choose n}{k\choose m}}$$

And the function \$g()\$ is used to compute each term \$a_i\$ for \$1\le i \le n+m\$.

#JavaScript (ES6), 111 bytes

This is really just a port of Delfad0r's great answer, implemented as nested recursive functions.

Takes input as (n)(m).

n=>m=>(g=i=>i?[...g(i-1),(h=k=>k&&h(k-1)-(b=(x,y)=>y?x*b(x-1,y-1)/y:i-k&1||-1)(i,k)*b(k,n)*b(k,m))(i)]:[])(n+m) 

Try it online!

###How?

The function \$b(x,y)\$ computes:

$$-(-1)^{i-k}{x\choose y}$$

It's slightly shorter to return \$-(-1)^{i-k}\$ as the last iteration of this function than computing it separately. Because we compute the product of three binomial coefficients, the signs of the first two of them are simply cancelling each other out.

The function \$h()\$ is used to compute:

$$\sum_{k=1}^{i}-b(i,k)\times b(k,n)\times b(k,m)\\ =\sum_{k=1}^{i}{(-1)^{i-k}{i\choose k}{k\choose n}{k\choose m}}$$

And the function \$g()\$ is used to compute each term \$a_i\$ for \$1\le i \le n+m\$.

#JavaScript (ES6), 109 bytes

This is really just a port of Delfad0r's great answer, implemented as nested recursive functions.

Takes input as (n)(m).

n=>m=>(g=i=>i?[...g(i-1),(h=k=>k&&h(k-1)-(b=(y,x=k)=>y?x*b(y-1,x-1)/y:i-k&1||-1)(k,i)*b(n)*b(m))(i)]:[])(n+m) 

Try it online!

###How?

The function \$b(y,x)\$ computes:

$$-(-1)^{i-k}{x\choose y}$$

It's slightly shorter to return \$-(-1)^{i-k}\$ as the last iteration of this function than processing it separately. Because we compute the product of three binomial coefficients, the signs of the first two of them are simply cancelling each other out.

The function \$h()\$ is used to compute:

$$\sum_{k=1}^{i}-b(k,i)\times b(n,k)\times b(m,k)\\ =\sum_{k=1}^{i}{(-1)^{i-k}{i\choose k}{k\choose n}{k\choose m}}$$

And the function \$g()\$ is used to compute each term \$a_i\$ for \$1\le i \le n+m\$.

#JavaScript (ES6), 111 bytes

This is really just a port of Delfad0r's great answer, implemented as nested recursive functions.

Takes input as (n)(m).

n=>m=>(g=i=>i?[...g(i-1),(h=k=>k&&h(k-1)-(b=(x,y)=>y?x*b(x-1,y-1)/y:i-k&1||-1)(i,k)*b(k,n)*b(k,m))(i)]:[])(n+m) 

Try it online!

#JavaScript (ES6), 111 bytes

This is really just a port of Delfad0r's great answer, implemented as nested recursive functions.

Takes input as (n)(m).

n=>m=>(g=i=>i?[...g(i-1),(h=k=>k&&h(k-1)-(b=(x,y)=>y?x*b(x-1,y-1)/y:i-k&1||-1)(i,k)*b(k,n)*b(k,m))(i)]:[])(n+m) 

Try it online!

###How?

The function \$b(x,y)\$ computes:

$$-(-1)^{i-k}{x\choose y}$$

It's slightly shorter to return \$-(-1)^{i-k}\$ as the last iteration of this function than computing it separately. Because we compute the product of three binomial coefficients, the signs of the first two of them are simply cancelling each other out.

The function \$h()\$ is used to compute:

$$\sum_{k=1}^{i}-b(i,k)\times b(k,n)\times b(k,m)\\ =\sum_{k=1}^{i}{(-1)^{i-k}{i\choose k}{k\choose n}{k\choose m}}$$

And the function \$g()\$ is used to compute each term \$a_i\$ for \$1\le i \le n+m\$.